問
original: https://www.youtube.com/watch?v=PvL2RDp1H4Y
x・exp(-α x^2)の広義積分[0,∞]
\(\int_0^{\infty} x e^{-\alpha x^2} \mathrm{d}x\)
$$\begin{eqnarray} &&\int_0^{\infty} x e^{-\alpha x^2} \mathrm{d}x \\&=&\int_0^{\infty} \sqrt{\frac{t}{\alpha}} e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&&\;\cdots\;t=\alpha x^2,\;x=0 \rightarrow t=0,\;x=\infty \rightarrow t=\infty, \\&&\;\cdots\;x=\sqrt{\frac{t}{\alpha}}\;(積分範囲からx\geq 0とする(?)) \\&&\;\cdots\;\frac{\mathrm{d}t}{\mathrm{d}x}=2\alpha x,\;\mathrm{d}x=\frac{1}{2\alpha x}\mathrm{d}t=\frac{1}{2\alpha \sqrt{\frac{t}{\alpha}}}\mathrm{d}t=\frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\int_0^{\infty}\left(\frac{t}{\alpha}\right)^{\frac{1}{2}} e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\frac{1}{2\alpha^{\frac{1}{2}}\sqrt{\alpha}}\int_0^{\infty} \frac{t^{\frac{1}{2}}}{\sqrt{t}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha}\int_0^{\infty} e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha}\int_0^{\infty} 1 \cdot e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha}\int_0^{\infty} t^0 \cdot e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha}\int_0^{\infty} t^{1-1} e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha}\Gamma\left(1\right)\;\cdots\;\Gamma(z)=\int_0^{\infty} t^{z-1} e^{-t}\mathrm{d}t\;(\mathfrak{Re}(z)>0) \\&=&\frac{1}{2\alpha}\;\left(1-1\right)!\;\cdots\;\Gamma(n)=(n-1)!\;(nは自然数) \\&=&\frac{1}{2\alpha}\cdot 1\;\cdots\;0!=1 \\&=&\frac{1}{2\alpha} \end{eqnarray}$$x^2・exp(-α x^2) の広義積分[0, ∞]
\(\int_0^{\infty} x^2 e^{-\alpha x^2} \mathrm{d}x\)
$$\begin{eqnarray} &&\int_0^{\infty} x^2 e^{-\alpha x^2} \mathrm{d}x \\&=&\int_0^{\infty} \frac{t}{\alpha} e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&&\;\cdots\;t=\alpha x^2,\;x=0 \rightarrow t=0,\;x=\infty \rightarrow t=\infty, \\&&\;\cdots\;x^2=\frac{t}{\alpha} \\&&\;\cdots\;x=\sqrt{\frac{t}{\alpha}}\;(積分範囲からx\geq 0とする(?)) \\&&\;\cdots\;\frac{\mathrm{d}t}{\mathrm{d}x}=2\alpha x,\;\mathrm{d}x=\frac{1}{2\alpha x}\mathrm{d}t=\frac{1}{2\alpha \sqrt{\frac{t}{\alpha}}}\mathrm{d}t=\frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\int_0^{\infty} \frac{t}{\alpha} e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\int_0^{\infty} \frac{t}{\sqrt{t}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\int_0^{\infty} t t^{-\frac{1}{2}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\int_0^{\infty} t^{\frac{1}{2}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\int_0^{\infty} t^{\frac{3}{2}-1}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\Gamma\left(\frac{3}{2}\right)\;\cdots\;\Gamma(z)=\int_0^{\infty} t^{z-1} e^{-t}\mathrm{d}t\;(\mathfrak{Re}(z)>0) \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\Gamma\left(\frac{1}{2}+1\right) \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\frac{(2\cdot 1-1)!!}{2^1}\sqrt{\pi}\;\cdots\;\Gamma\left(\frac{1}{2}+n\right)=\frac{(2n-1)!!}{2^n}\sqrt{\pi}\;(!!は二重階乗) \;(nは自然数) \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\frac{1!!}{2}\sqrt{\pi} \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\frac{1}{2}\sqrt{\pi}\;\cdots\;nの二重階乗は，1 から n まで n と同じ偶奇性を持つものだけを全て掛けた積 \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\frac{1\sqrt{\pi}}{2} \\&=&\frac{1}{4\alpha}\sqrt{\frac{\pi}{\alpha}} \end{eqnarray}$$x^3・exp(-α x^2)の広義積分[0,∞]
\(\int_0^{\infty} x^3 e^{-\alpha x^2} \mathrm{d}x\)
$$\begin{eqnarray} &&\int_0^{\infty} x^3 e^{-\alpha x^2} \mathrm{d}x \\&=&\int_0^{\infty} \left(\frac{t}{\alpha}\right)^2 e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&&\;\cdots\;t=\alpha x^2,\;x=0 \rightarrow t=0,\;x=\infty \rightarrow t=\infty, \\&&\;\cdots\;x=\sqrt{\frac{t}{\alpha}}\;(積分範囲からx\geq 0とする(?)) \\&&\;\cdots\;x^3=\left(\sqrt{\frac{t}{\alpha}}\right)^3=\left(\frac{t}{\alpha}\right)^{\frac{3}{2}} \\&&\;\cdots\;\frac{\mathrm{d}t}{\mathrm{d}x}=2\alpha x,\;\mathrm{d}x=\frac{1}{2\alpha x}\mathrm{d}t=\frac{1}{2\alpha \sqrt{\frac{t}{\alpha}}}\mathrm{d}t=\frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\int_0^{\infty}\left(\frac{t}{\alpha}\right)^{\frac{3}{2}} e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\frac{1}{2\alpha^{\frac{3}{2}}\sqrt{\alpha}}\int_0^{\infty} \frac{t^{\frac{3}{2}}}{\sqrt{t}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha^2}\int_0^{\infty} t e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha^2}\int_0^{\infty} t^{2-1} e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha^2}\Gamma\left(2\right)\;\cdots\;\Gamma(z)=\int_0^{\infty} t^{z-1} e^{-t}\mathrm{d}t\;(\mathfrak{Re}(z)>0) \\&=&\frac{1}{2\alpha^2}\;\left(2-1\right)!\;\cdots\;\Gamma(n)=(n-1)!\;(nは自然数) \\&=&\frac{1}{2\alpha^2}\cdot 1 \\&=&\frac{1}{2\alpha^2} \end{eqnarray}$$x^4・exp(-α x^2) の広義積分[0,∞]
\(\int_0^{\infty} x^4 e^{-\alpha x^2} \mathrm{d}x\)
$$\begin{eqnarray} &&\int_0^{\infty} x^4 e^{-\alpha x^2} \mathrm{d}x \\&=&\int_0^{\infty} \left(\frac{t}{\alpha}\right)^2 e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&&\;\cdots\;t=\alpha x^2,\;x=0 \rightarrow t=0,\;x=\infty \rightarrow t=\infty, \\&&\;\cdots\;x^2=\frac{t}{\alpha},\;x^4=x^{2^2}=\left(\frac{t}{\alpha}\right)^2 \\&&\;\cdots\;x=\sqrt{\frac{t}{\alpha}}\;(積分範囲からx\geq 0とする(?)) \\&&\;\cdots\;\frac{\mathrm{d}t}{\mathrm{d}x}=2\alpha x,\;\mathrm{d}x=\frac{1}{2\alpha x}\mathrm{d}t=\frac{1}{2\alpha \sqrt{\frac{t}{\alpha}}}\mathrm{d}t=\frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\int_0^{\infty} \left(\frac{t}{\alpha}\right)^2 e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\int_0^{\infty} \frac{t^2}{\sqrt{t}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\int_0^{\infty} t^2 t^{-\frac{1}{2}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\int_0^{\infty} t^{\frac{3}{2}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\int_0^{\infty} t^{\frac{5}{2}-1}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\Gamma\left(\frac{5}{2}\right)\;\cdots\;\Gamma(z)=\int_0^{\infty} t^{z-1} e^{-t}\mathrm{d}t\;(\mathfrak{Re}(z)>0) \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\Gamma\left(\frac{1}{2}+2\right) \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\frac{(2\cdot 2-1)!!}{2^2}\sqrt{\pi}\;\cdots\;\Gamma\left(\frac{1}{2}+n\right)=\frac{(2n-1)!!}{2^n}\sqrt{\pi}\;(!!は二重階乗) \;(nは自然数) \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\frac{3!!}{4}\sqrt{\pi} \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\frac{3\cdot 1}{4}\sqrt{\pi}\;\cdots\;nの二重階乗は，1 から n まで n と同じ偶奇性を持つものだけを全て掛けた積 \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\frac{3\sqrt{\pi}}{4} \\&=&\frac{3}{8\alpha^2}\sqrt{\frac{\pi}{\alpha}} \end{eqnarray}$$ガンマ凾数の相反公式の積分表示
ガンマ凾数の相反公式の積分表示
ベータ凾数のガンマ凾数での表示
$$\begin{eqnarray} \left.B\left(p,q\right)\right|_{p=1-q}&=&\href{https://shikitenkai.blogspot.com/2020/05/blog-post_22.html}{\left.\frac{\Gamma\left(p\right)\Gamma\left(q\right)}{\Gamma\left(p+q\right)}\right|_{p=1-q}} \;\ldots\;p,q\in\mathbb{C},\;\Re\left(p\right),\Re\left(q\right)\gt0 \\&=&\frac{\Gamma\left(1-q\right)\Gamma\left(q\right)}{\Gamma\left(1-q+q\right)} \\&=&\frac{\Gamma\left(1-q\right)\Gamma\left(q\right)}{\Gamma\left(1\right)} \\&=&\frac{\Gamma\left(1-q\right)\Gamma\left(q\right)}{1} \\&&\;\ldots\;\Gamma\left(1\right)=\int_0^\infty t^{1-1}e^{-t}\mathrm{d}t=\int_0^\infty t^{0}e^{-t}\mathrm{d}t=\int_0^\infty 1\cdot e^{-t}\mathrm{d}t=\int_0^\infty e^{-t}\mathrm{d}t \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/e-ax0.html}{\int_0^\infty e^{-t}\mathrm{d}t=\frac{1}{1}}=1 \\&=&\Gamma\left(1-q\right)\Gamma\left(q\right) \end{eqnarray}$$ベータ凾数の定義での表示
$$\begin{eqnarray} \left.B\left(p,q\right)\right|_{p=1-q} &=&\int_0^1 t^{p-1}\left(1-t\right)^{q-1}\mathrm{d}t \\&=&\int_\infty^0 \left(\frac{1}{x+1}\right)^{p-1}\left(1-\frac{1}{x+1}\right)^{q-1}\cdot\frac{-1}{\left(x+1\right)^2}\mathrm{d}x \\&&\;\ldots\;t=\frac{1}{x+1},\;x=\frac{1}{t}-1,\;\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{-1}{\left(x+1\right)^2} \\&&\;\ldots\;t:0\rightarrow1,\;x:\infty\rightarrow0 \\&=&\int_0^\infty \left(\frac{1}{x+1}\right)^{p-1}\left(1-\frac{1}{x+1}\right)^{q-1}\cdot\frac{1}{\left(x+1\right)^2}\mathrm{d}x \\&=&\int_0^\infty \left(\frac{1}{x+1}\right)^{p-1}\left(\frac{x}{x+1}\right)^{q-1}\cdot\frac{1}{\left(x+1\right)^2}\mathrm{d}x \\&=&\int_0^\infty \frac{1^{p-1}\cdot x^{q-1}\cdot 1}{\left(x+1\right)^{p-1}\left(x+1\right)^{q-1}\left(x+1\right)^{2}}\mathrm{d}x \\&=&\int_0^\infty \frac{x^{q-1}}{\left(x+1\right)^{(p-1)+(q-1)+2}}\mathrm{d}x \\&=&\int_0^\infty \frac{x^{q-1}}{\left(x+1\right)^{p+q}}\mathrm{d}x \\&=&\int_0^\infty \frac{x^{q-1}}{\left(x+1\right)^{1-q+q}}\mathrm{d}x \\&=&\int_0^\infty \frac{x^{q-1}}{x+1}\mathrm{d}x \end{eqnarray}$$ガンマ凾数の相反公式の積分表示
$$\begin{eqnarray} \Gamma\left(1-q\right)\Gamma\left(q\right)&=&\int_0^\infty \frac{x^{q-1}}{x+1}\mathrm{d}x \end{eqnarray}$$xe^(-ax)の広義積分[0,∞]
\(xe^{-ax}\)の広義積分\([0,\infty]\)
\begin{eqnarray} \int_0^\infty xe^{-ax} \mathrm{d}x &=&\lim_{n\rightarrow\infty} \int_0^n xe^{-ax} \mathrm{d}x \\&=&\lim_{n\rightarrow\infty} \left[ \left[x \cdot \frac{-1}{a}e^{-ax} \right]_0^n - \int_0^n \frac{-1}{a} \cdot e^{-ax} \mathrm{d}x \right] \\&=&\lim_{n\rightarrow\infty} \left[ \left[n \frac{-1}{a}e^{-an} - 0 \cdot \frac{-1}{a}e^{-a 0}\right] + \frac{1}{a} \int_0^n e^{-ax} \mathrm{d}x \right] \\&=&\lim_{n\rightarrow\infty} \left[ n \frac{-1}{a}e^{-an} + \frac{1}{a} \int_0^n \cdot e^{-ax} \mathrm{d}x \right] \\&=&\frac{-1}{a}\lim_{n\rightarrow\infty} ne^{-an} + \frac{1}{a} \lim_{n\rightarrow\infty} \int_0^n \cdot e^{-ax} \mathrm{d}x \\&=&0 + \frac{1}{a}\lim_{n\rightarrow\infty} \int_0^n e^{-ax} \mathrm{d}x \\&&指数凾数\left(e^{-x}\right)の方が線形凾数\left(x\right)より早く漸近化するので\lim_{n\rightarrow\infty}e^{-ax}=0が収束先となる． \\&=&\frac{1}{a}\lim_{n\rightarrow\infty} \int_0^n e^{-ax} \mathrm{d}x \\&=&\frac{1}{a}\cdot\frac{1}{a}\;\ldots\;\int_0^\infty e^{-ax} \mathrm{d}x=\frac{1}{a} \\&=&\frac{1}{a^2} \end{eqnarray}e^(-ax)の広義積分[0,∞]
\(e^{-ax}\)の広義積分\([0,\infty]\)
\begin{eqnarray} \int_0^\infty e^{-ax} \mathrm{d}x &=&\lim_{n\rightarrow\infty} \int_0^n e^{-ax} \mathrm{d}x \\&=&\lim_{n\rightarrow\infty} \int_0^{-\infty} e^{u} \left(\frac{-1}{a}\right)\mathrm{d}u \\&&\;\ldots\;u=-ax,\;\frac{\mathrm{d}u}{\mathrm{d}x}=-a,\;\mathrm{d}x=\frac{-1}{a}\mathrm{d}u \\&&\;\ldots\;x:0\rightarrow n,\;u:0\rightarrow -an \\&=&\lim_{n\rightarrow\infty} \frac{-1}{a}\int_0^{-an} e^{u}\mathrm{d}u \\&=&\lim_{n\rightarrow\infty} \frac{-1}{a}\left[e^{u} \right]_0^{-an} \\&=&\lim_{n\rightarrow\infty} \frac{-1}{a}\left[e^{-an} - e^{0}\right] \\&=&\frac{-1}{a}\left[0 - 1\right] \\&=&\frac{-1}{a}\cdot(-1) \\&=&\frac{1}{a} \end{eqnarray}u^(z-1)/(1-u)における定積分[0,1],もしくは1/(z+k)の無限級数
\(\frac{u^{z-1}}{1-u}\)における定積分[0,1],もしくは\(\frac{1}{z+k}\)の無限級数
$$\begin{eqnarray} \int_{0}^1 \frac{u^{z-1}}{1-u} \mathrm{d}u &=&\int_{0}^1 u^{z-1}\frac{1}{1-u} \mathrm{d}u \\&=&\int_{0}^1 u^{z-1}\left(1+u+u^2+\cdots\right) \mathrm{d}u \\&&\;\ldots\href{https://shikitenkai.blogspot.com/2021/07/11-x-1.html}{\frac{1}{1-u}=1+u+u^2+\cdots\;\;\;\left(|u|\lt1\right)} \\&=&\int_{0}^1 \left(u^{z-1}+u^{z-1}u+u^{z-1}u^2+\cdots\right) \mathrm{d}u \\&=&\int_{0}^1 \left(u^{z-1}+u^{z-1+1}+u^{z-1+2}+\cdots\right) \mathrm{d}u \\&=&\int_{0}^1 \left(u^{z-1}+u^{z}+u^{z+1}+\cdots\right) \mathrm{d}u \\&=&\left[\frac{1}{z-1+1}u^{z-1+1}+\frac{1}{z+1}u^{z+1}+\frac{1}{z+1+1}u^{z+1+1}+\cdots\right]_0^1 \\&=&\left[\frac{1}{z}u^z+\frac{1}{z+1}u^{z+1}+\frac{1}{z+2}u^{z+2}+\cdots\right]_0^1 \\&=&\left[\frac{1}{z}1^z-\frac{1}{z}0^z+\frac{1}{z+1}1^{z+1}-\frac{1}{z+1}0^{z+1}+\frac{1}{z+2}1^{z+2}-\frac{1}{z+2}0^{z+2}+\cdots\right] \\&=&\frac{1}{z}+\frac{1}{z+1}+\frac{1}{z+2}+\cdots \\&=&\sum_{k=0}^\infty \frac{1}{z+k} \end{eqnarray}$$1/(x^2+px-1)の不定積分
1/(x^2+px-1)の不定積分
$$\begin{eqnarray} \int \frac{1}{x^2+px-1}\mathrm{d}x\;\ldots\;pは実数 \end{eqnarray}$$1/(x^2+px-1)の部分分数分解
$$\begin{eqnarray} \frac{1}{x^2+px-1}&=&\frac{1}{x^2+\color{red}{2}\color{black}{}\frac{p}{\color{red}{2}\color{black}{}}x \color{blue}{+\left(\frac{p}{2}\right)^2-\left(\frac{p}{2}\right)^2}\color{black}{}-1} \\&=&\frac{1}{\left(x+\frac{p}{2}\right)^2-\left(\frac{p^2}{4}+1\right)} \\&=&\frac{1}{\left(x+\frac{p}{2}\right)^2-\left(\sqrt{\frac{p^2}{4}+1}\right)^2} \;\ldots\;\frac{p^2}{4}+1\gt0なので問題なく根号をつけられる \\&=&\frac{1}{\left(x+\frac{p}{2}+\sqrt{\frac{p^2}{4}+1}\right)\left(x+\frac{p}{2}-\sqrt{\frac{p^2}{4}+1}\right)} \\&=&\frac{1}{\lambda_1\lambda_2} \;\ldots\;\lambda_{1,2}=x+\frac{p}{2}\pm\sqrt{\frac{p^2}{4}+1} \\&=&\frac{C_1}{\lambda_1}+\frac{C_2}{\lambda_2}=\frac{C_1\lambda_2+C_2\lambda_1}{\lambda_1\lambda_2} \;\ldots\;部分分数分解 \\&=&\frac{\frac{-1}{\sqrt{p^2+4}}}{\lambda_1} +\frac{\frac{1}{\sqrt{p^2+4}}}{\lambda_2} \\&&\;\ldots\;\small{C_1\lambda_2+C_2\lambda_1=1} \\&&\;\ldots\;\small{C_1 x+C_1\left(\frac{p}{2}-\sqrt{\frac{p^2}{4}+1}\right)+C_2x+C_2\left(\frac{p}{2}+\sqrt{\frac{p^2}{4}+1}\right)=1} \\&&\;\ldots\;\left\{ \begin{array} \\(C_1+C_2)x=0 \\C_1\left(\frac{p}{2}-\sqrt{\frac{p^2}{4}+1}\right)+C_2\left(\frac{p}{2}+\sqrt{\frac{p^2}{4}+1}\right)=1 \end{array} \right.\;\ldots\;係数比較 \\&&\;\ldots\;C_1=-C_2\;\ldots\;一つ目の式より \\&&\;\ldots\;C_1\left(\frac{p}{2}-\sqrt{\frac{p^2}{4}+1}\right)-C_1\left(\frac{p}{2}+\sqrt{\frac{p^2}{4}+1}\right)=1 \\&&\;\ldots\;C_1\left\{\left(\frac{p}{2}-\sqrt{\frac{p^2}{4}+1}\right)-\left(\frac{p}{2}+\sqrt{\frac{p^2}{4}+1}\right)\right\}=1 \\&&\;\ldots\;-2 C_1 \sqrt{\frac{p^2}{4}+1}=1 \\&&\;\ldots\;-C_1 \sqrt{p^2+4}=1 \\&&\;\ldots\;C_1 =\frac{-1}{\sqrt{p^2+4}} \\&&\;\ldots\;C_2 =-C_1=\frac{1}{\sqrt{p^2+4}} \\&=&\frac{1}{\sqrt{p^2+4}}\left(\frac{-1}{\lambda_1}+\frac{1}{\lambda_2}\right) \\&=&\frac{1}{\sqrt{p^2+4}}\left(\frac{-1}{x+\frac{p}{2}+\sqrt{\frac{p^2}{4}+1}}+\frac{1}{x+\frac{p}{2}-\sqrt{\frac{p^2}{4}+1}}\right) \end{eqnarray}$$1/(x^2+px-1)の不定積分 計算
$$\begin{eqnarray} \int \frac{1}{x^2+px-1}\mathrm{d}x &=&\int \frac{1}{\sqrt{p^2+4}}\left(\frac{-1}{x+\frac{p}{2}+\sqrt{\frac{p^2}{4}+1}} +\frac{1}{x+\frac{p}{2}-\sqrt{\frac{p^2}{4}+1}}\right)\mathrm{d}x \\&=&\frac{1}{\sqrt{p^2+4}}\left( \int \frac{-1}{x+\frac{p}{2}+\sqrt{\frac{p^2}{4}+1}} \mathrm{d}x +\int \frac{1}{x+\frac{p}{2}-\sqrt{\frac{p^2}{4}+1}} \mathrm{d}x \right) \\&=&\frac{1}{\sqrt{p^2+4}}\left( \int \frac{-1}{u} \mathrm{d}u +\int \frac{1}{v} \mathrm{d}v \right) \\&&\;\ldots\;u=x+\frac{p}{2}+\sqrt{\frac{p^2}{4}+1},\;\frac{\mathrm{d}u}{\mathrm{d}x}=1,\;\mathrm{d}x=\mathrm{d}u \\&&\;\ldots\;v=x+\frac{p}{2}-\sqrt{\frac{p^2}{4}+1},\;\frac{\mathrm{d}v}{\mathrm{d}x}=1,\;\mathrm{d}x=\mathrm{d}v \\&=&\frac{1}{\sqrt{p^2+4}}\left( -\log{\left|u\right|} +\log{\left|v\right|} \right) \\&&\;\ldots\;\int \frac{1}{x} \mathrm{d}x=\log{\left|x\right|}+C\;\;(C:積分定数) \\&=&\frac{1}{\sqrt{p^2+4}}\left( -\log{\left|x+\frac{p}{2}+\sqrt{\frac{p^2}{4}+1}\right|} +\log{\left|x+\frac{p}{2}-\sqrt{\frac{p^2}{4}+1}\right|} \right) \\&=&\frac{1}{\sqrt{p^2+4}}\log{\frac{ \left|x+\frac{p}{2}-\sqrt{\frac{p^2}{4}+1}\right| }{ \left|x+\frac{p}{2}+\sqrt{\frac{p^2}{4}+1}\right| }} + C\;\;(C:積分定数) \\&&\;\ldots\;\log{A}-\log{B}=\log{\frac{A}{B}} \end{eqnarray}$$\(p=0\)の場合
$$\begin{eqnarray} \left.\int \frac{1}{x^2+px-1}\mathrm{d}x\right|_{p=0} &=&\left.\frac{1}{\sqrt{p^2+4}}\log{\frac{ \left|x+\frac{p}{2}-\sqrt{\frac{p^2}{4}+1}\right| }{ \left|x+\frac{p}{2}+\sqrt{\frac{p^2}{4}+1}\right| }}\right|_{p=0} \\&=&\frac{1}{\sqrt{0^2+4}}\log{\frac{ \left|x+\frac{0}{2}-\sqrt{\frac{0^2}{4}+1}\right| }{ \left|x+\frac{0}{2}+\sqrt{\frac{0^2}{4}+1}\right| }} \\&=&\frac{1}{2}\log{\frac{ \left|x-1\right| }{ \left|x+1\right| }} + C\;\;(C:積分定数) \end{eqnarray}$$定積分を用いた凾数同士の被積分凾数
定積分を用いた凾数同士の被積分凾数
$$ \begin{eqnarray} F(x)&=&\int_a^x f(t)\mathrm{d}t\;\cdots\;定積分を用いた凾数 \\G(x)&=&\int_a^x g(t)\mathrm{d}t \\ただし,&&a\leq x \leq b \end{eqnarray} $$ とする．今，任意の\(x\)において(\(x\)の取りえる範囲すべての\(x\)において) $$ \begin{eqnarray} F(x)&=&G(x) \end{eqnarray} $$ が成り立つなら(特定の積分範囲の定積分結果が等しいだけではないところに注意)， $$ \begin{eqnarray} F^\prime(x)&=&G^\prime(x) \end{eqnarray} $$ つまり $$ \begin{eqnarray} f(x)&=&g(x) \end{eqnarray} $$ が成り立つ．よって $$ \begin{eqnarray} \int_a^x f(t)\mathrm{d}t &=&\int_a^x g(t)\mathrm{d}t \end{eqnarray} $$ ならば $$ \begin{eqnarray} \\f(x)&=&g(x) \end{eqnarray} $$ であり，被積分凾数同士も等しい．
x/√(a-x)の積分
x/√(a-x)の積分
不定積分
$$ \begin{eqnarray} \int \frac{x}{\sqrt{a-x}}\mathrm{d}x &=&\int x(a-x)^{-\frac{1}{2}}\mathrm{d}x \\&=&\int x\left\{-2(a-x)^{\frac{1}{2}}\right\}^\prime \mathrm{d}x \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/1a-x.html}{\int (a-x)^{-\frac{1}{2}} \mathrm{d}x=-2(a-x)^{\frac{1}{2}}+C} \\&=&-2x(a-x)^{\frac{1}{2}}-\int -2(a-x)^{\frac{1}{2}} \mathrm{d}x \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/02/blog-post_7.html}{\int f^\prime(x)g(x) \mathrm{d}x= fg-\int fg^\prime \mathrm{d}x} \\&=&-2x(a-x)^{\frac{1}{2}}+2\int (a-x)^{\frac{1}{2}} \mathrm{d}x \\&=&-2x(a-x)^{\frac{1}{2}}+2\left\{\frac{1}{\frac{1}{2}+1}(a-x)^{\frac{1}{2}+1}(-1)\right\} \\&=&-2x(a-x)^{\frac{1}{2}}+2\left\{\frac{-1}{\frac{3}{2}}(a-x)^{\frac{3}{2}}\right\} \\&=&-2x(a-x)^{\frac{1}{2}}+2\left\{\frac{-2}{3}(a-x)^{\frac{3}{2}}\right\} \\&=&-2x(a-x)^{\frac{1}{2}}-\frac{4}{3}(a-x)^{\frac{3}{2}} \\&=&-2x(a-x)^{\frac{1}{2}}-\frac{4}{3}(a-x)(a-x)^{\frac{1}{2}} \\&=&(a-x)^{\frac{1}{2}}\left\{-2x-\frac{4}{3}(a-x)\right\} \\&=&(a-x)^{\frac{1}{2}}\left(-2x-\frac{4}{3}a+\frac{4}{3}x\right) \\&=&(a-x)^{\frac{1}{2}}\left(-\frac{6}{3}x-\frac{4}{3}a+\frac{4}{3}x\right) \\&=&(a-x)^{\frac{1}{2}}\left(-\frac{2}{3}x-\frac{4}{3}a\right) \\&=&-\frac{2}{3}(a-x)^{\frac{1}{2}}\left(x+2a\right)+C\;\cdots\;C:積分定数 \\&=&-\frac{2}{3}\sqrt{a-x}\left(x+2a\right)+C \end{eqnarray} $$定積分
$$ \begin{eqnarray} \int_{0}^{a} \frac{x}{\sqrt{a-x}}\mathrm{d}x &=&\left[-\frac{2}{3}\sqrt{a-x}\left(x+2a\right)\right]_{0}^{a} \\&=&(-\frac{2}{3}\sqrt{a-a}\left(a+2a\right))-(-\frac{2}{3}\sqrt{a-0}\left(0+2a\right)) \\&=&0-(-\frac{2}{3}\sqrt{a}\left(2a\right)) \\&=&0-(-\frac{4}{3}\sqrt{a}^3) \\&=&\frac{4}{3}a^{\frac{3}{2}} \end{eqnarray} $$√(x)/√(a-x)の積分
√(x)/√(a-x)の積分
不定積分
$$
\begin{eqnarray}
\int \frac{\sqrt{x}}{\sqrt{a-x}}\mathrm{d}x
&=&\int x^{\frac{1}{2}}(a-x)^{-\frac{1}{2}}\mathrm{d}x
\\&=&\int u(a-u^2)^{-\frac{1}{2}}2u\mathrm{d}u
\;\cdots\;u=\sqrt{x},\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{1}{2\sqrt{x}},\mathrm{d}x=2\sqrt{x}\mathrm{d}u=2u\mathrm{d}u
\\&=&2\int u^2(a-u^2)^{-\frac{1}{2}}\mathrm{d}u
\;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x
\\&=&2\int \left\{\sqrt{a}\sin{\left(\theta\right)}\right\}^2\left[a-\left\{\sqrt{a}\sin{\left(\theta\right)}\right\}^2\right]^{-\frac{1}{2}}\sqrt{a}\cos{\left(\theta\right)}\mathrm{d}\theta
\;\cdots\;u=\sqrt{a}\sin{\left(\theta\right)},\frac{\mathrm{d}u}{\mathrm{d}\theta}=\sqrt{a}\cos{\left(\theta\right)},\mathrm{d}u=\sqrt{a}\cos{\left(\theta\right)}\mathrm{d}\theta
\\&=&2\int a\sin^2{\left(\theta\right)}\left\{a-a\sin^2{\left(\theta\right)}\right\}^{-\frac{1}{2}}\sqrt{a}\cos{\left(\theta\right)}\mathrm{d}\theta
\\&=&2a\sqrt{a}\int \sin^2{\left(\theta\right)}\left[a\left\{1-\sin^2{\left(\theta\right)}\right\}\right]^{-\frac{1}{2}}\cos{\left(\theta\right)}\mathrm{d}\theta
\\&=&2a\sqrt{a}\int \sqrt{a}\sin^2{\left(\theta\right)}a^{-\frac{1}{2}}\left\{1-\sin^2{\left(\theta\right)}\right\}^{-\frac{1}{2}}\cos{\left(\theta\right)}\mathrm{d}\theta
\\&=&2a\sqrt{a}a^{-\frac{1}{2}}\int \sin^2{\left(\theta\right)}\left\{1-\sin^2{\left(\theta\right)}\right\}^{-\frac{1}{2}}\cos{\left(\theta\right)}\mathrm{d}\theta
\\&=&2a\int \sin^2{\left(\theta\right)}\left\{1-\sin^2{\left(\theta\right)}\right\}^{-\frac{1}{2}}\cos{\left(\theta\right)}\mathrm{d}\theta
\\&=&2a\int \sin^2{\left(\theta\right)}\left\{\cos^2{\left(\theta\right)}\right\}^{-\frac{1}{2}}\cos{\left(\theta\right)}\mathrm{d}\theta
\\&=&2a\int \sin^2{\left(\theta\right)}\left\{\cos{\left(\theta\right)}\right\}^{-1}\cos{\left(\theta\right)}\mathrm{d}\theta
\\&=&2a\int \sin^2{\left(\theta\right)}\mathrm{d}\theta
\\&=&2a\left\{\frac{1}{2}\theta-\frac{1}{2}\sin{\left(\theta\right)}\cos{\left(\theta\right)}\right\}
\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/sin.html}{\int \sin^2{\left(\theta\right)}\mathrm{d}\theta=\frac{1}{2}\left\{\theta-\sin{\left(\theta\right)}\cos{\left(\theta\right)}\right\}+C\;(C:積分定数)}
\\&=&2a\left\{\frac{1}{2}\sin^{-1}{\left(\frac{u}{\sqrt{a}}\right)}-\frac{1}{2}\sin{\left(\sin^{-1}{\left(\frac{u}{\sqrt{a}}\right)} \right)}\cos{\left(\sin^{-1}{\left(\frac{u}{\sqrt{a}}\right)} \right)}\right\}
\;\cdots\;u=\sqrt{a}\sin{\left(\theta\right)},\;\frac{u}{\sqrt{a}}=\sin{\left(\theta\right)},\;\theta=\sin^{-1}{\left(\frac{u}{\sqrt{a}}\right)}
\\&=&2a\left\{\frac{1}{2}\sin^{-1}{\left(\frac{u}{\sqrt{a}}\right)}-\frac{1}{2}\left(\frac{u}{\sqrt{a}}\right)\sqrt{ 1-\left(\frac{u}{\sqrt{a}}\right)^2 }\right\}
\;\cdots\;\sin{\left(\sin^{-1}{\left(x\right)}\right)}=x,\;\cos{\left(\sin^{-1}{\left(x\right)}\right)}=\sqrt{1-x^2}
\\&=&2a\frac{1}{2}\left\{\sin^{-1}{\left(\frac{u}{\sqrt{a}}\right)}-\left(\frac{u}{\sqrt{a}}\right)\sqrt{ 1-\left(\frac{u}{\sqrt{a}}\right)^2 }\right\}
\\&=&a\left\{\sin^{-1}{\left(\frac{u}{\sqrt{a}}\right)}-\left(\frac{u}{\sqrt{a}}\right)\sqrt{ 1-\left(\frac{u}{\sqrt{a}}\right)^2 }\right\}
\\&=&a\left\{\sin^{-1}{\left(\frac{\sqrt{x}}{\sqrt{a}}\right)}-\left(\frac{\sqrt{x}}{\sqrt{a}}\right)\sqrt{ 1-\left(\frac{\sqrt{x}}{\sqrt{a}}\right)^2 }\right\}
\;\cdots\;u=\sqrt{x}
\\&=&a\left\{\sin^{-1}{\left(\sqrt{\frac{x}{a}}\right)}-\frac{\sqrt{x}}{\sqrt{a}}\sqrt{ 1-\frac{x}{a} }\right\}
\\&=&a\left\{\sin^{-1}{\left(\sqrt{\frac{x}{a}}\right)}-\frac{\sqrt{x}}{\sqrt{a}}\sqrt{ \frac{a-x}{a} }\right\}
\\&=&a\left\{\sin^{-1}{\left(\sqrt{\frac{x}{a}}\right)}-\frac{\sqrt{x}}{\sqrt{a}}\frac{\sqrt{a-x}}{\sqrt{a}}\right\}
\;\cdots\;\sqrt{\frac{A}{B}}=\left(\frac{A}{B}\right)^{\frac{1}{2}}=\left(A\frac{1}{B}\right)^{\frac{1}{2}}=A^{\frac{1}{2}}\left(\frac{1}{B}\right)^{\frac{1}{2}}=A^{\frac{1}{2}}\left(B^{-1}\right)^{\frac{1}{2}}=A^{\frac{1}{2}}B^{-\frac{1}{2}}=\sqrt{A}\frac{1}{\sqrt{B}}=\frac{\sqrt{A}}{\sqrt{B}}
\\&=&a\left\{\sin^{-1}{\left(\sqrt{\frac{x}{a}}\right)}-\frac{\sqrt{x}\sqrt{a-x}}{a }\right\}
\\&=&a\sin^{-1}{\left(\sqrt{\frac{x}{a}}\right)}-a\frac{\sqrt{x}\sqrt{a-x}}{a}
\\&=&a\sin^{-1}{\left(\sqrt{\frac{x}{a}}\right)}-\sqrt{x}\sqrt{a-x}+C\;\cdots\;C:積分定数
\\&=&a\sin^{-1}{\left(\left(\frac{x}{a}\right)^{\frac{1}{2}}\right)}-x^{\frac{1}{2}}\left(a-x\right)^{\frac{1}{2}}+C
\end{eqnarray}
$$
定積分
$$
\begin{eqnarray}
\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{a-x}}\mathrm{d}x
&=&\left[a\sin^{-1}{\left(\left(\frac{x}{a}\right)^{\frac{1}{2}}\right)}-x^{\frac{1}{2}}\left(a-x\right)^{\frac{1}{2}}\right]_{0}^{a}
\\&=&(a\sin^{-1}{\left(\left(\frac{a}{a}\right)^{\frac{1}{2}}\right)}-a^{\frac{1}{2}}\left(a-a\right)^{\frac{1}{2}})-(a\sin^{-1}{\left(\left(\frac{0}{a}\right)^{\frac{1}{2}}\right)}-0^{\frac{1}{2}}\left(a-0\right)^{\frac{1}{2}})
\\&=&(a\sin^{-1}{\left(1\right)}-0)-(0-0)
\;\cdots\;\sin^{-1}{\left(0\right)}=0,\;0^A=0
\\&=&a \frac{\pi}{2}
\;\cdots\;\sin^{-1}{\left(1\right)}=\frac{\pi}{2}
\\&=&\frac{a\pi}{2}
\end{eqnarray}
$$
1/√(a-x)の積分
1/√(a-x)の積分
不定積分
$$ \begin{eqnarray} \int \frac{1}{\sqrt{a-x}}\mathrm{d}x &=&\int (a-x)^{-\frac{1}{2}}\mathrm{d}x \\&=&\int u^{-\frac{1}{2}}(-1)\mathrm{d}u \;\cdots\;u=a-x,\frac{\mathrm{d}u}{\mathrm{d}x}=-1,\mathrm{d}x=-\mathrm{d}u \\&=&-\int u^{-\frac{1}{2}}\mathrm{d}u \;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&-\frac{1}{-\frac{1}{2}+1}u^{-\frac{1}{2}+1} \;\cdots\;\int x^a \mathrm{d}x=\frac{1}{a+1}x^{a+1}+C\;(C:積分定数) \\&=&-\frac{1}{\frac{1}{2}}u^{\frac{1}{2}} \\&=&-2u^{\frac{1}{2}} \\&=&-2(a-x)^{\frac{1}{2}}\;\cdots\;u=a-x \\&=&-2(a-x)^{\frac{1}{2}}+C\;\cdots\;C:積分定数 \\&=&-2\sqrt{a-x}+C \end{eqnarray} $$定積分
$$ \begin{eqnarray} \int_{0}^{a} \frac{1}{\sqrt{a-x}}\mathrm{d}x &=&\left[-2\sqrt{a-x}\right]_{0}^{a} \\&=&(-2\sqrt{a-a})-(-2\sqrt{a-0}) \\&=&0-(-2\sqrt{a}) \\&=&2\sqrt{a} \\&=&2a^{\frac{1}{2}} \end{eqnarray} $$1/√(x(a-x))の積分
1/√(x(a-x))の積分
不定積分
$$ \begin{eqnarray} \int \frac{1}{\sqrt{x(a-x)}}\mathrm{d}x &=&\int \left\{x(a-x)\right\}^{-\frac{1}{2}}\mathrm{d}x \\&=&\int x^{-\frac{1}{2}}\left(a-x\right)^{-\frac{1}{2}}\mathrm{d}x \\&=&\int u^{-1}(a-u^2)^{-\frac{1}{2}}\;2u\mathrm{d}u \;\cdots\;u=\sqrt{x},\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{1}{2\sqrt{x}},\mathrm{d}x=2\sqrt{x}\mathrm{d}u=2u\mathrm{d}u \\&=&2\int (a-u^2)^{-\frac{1}{2}}\mathrm{d}u \;\cdots\;u^{-1}u=1,\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&2\int \left[a-\left\{\sqrt{a}\sin{\left(\theta\right)}\right\}^2\right]^{-\frac{1}{2}}\;\sqrt{a}\cos{\left(\theta\right)}\mathrm{d}\theta \;\cdots\;u=\sqrt{a}\sin{\left(\theta\right)},\frac{\mathrm{d}u}{\mathrm{d}\theta}=\sqrt{a}\cos{\left(\theta\right)},\mathrm{d}u=\sqrt{a}\cos{\left(\theta\right)}\mathrm{d}\theta \\&=&2\sqrt{a}\int \left[a-\left\{\sqrt{a}^2\sin^2{\left(\theta\right)}\right\}\right]^{-\frac{1}{2}}\;\cos{\left(\theta\right)}\mathrm{d}\theta \;\cdots\;(AB)^C=A^CB^C \\&=&2\sqrt{a}\int \left\{a-a\sin^2{\left(\theta\right)}\right\}^{-\frac{1}{2}}\;\cos{\left(\theta\right)}\mathrm{d}\theta \\&=&2\sqrt{a}\int \left[a\left\{1-\sin^2{\left(\theta\right)}\right\}\right]^{-\frac{1}{2}}\;\cos{\left(\theta\right)}\mathrm{d}\theta \\&=&2\sqrt{a}\int a^{-\frac{1}{2}}\left\{1-\sin^2{\left(\theta\right)}\right\}^{-\frac{1}{2}}\;\cos{\left(\theta\right)}\mathrm{d}\theta \\&=&2a\sqrt{a}a^{-\frac{1}{2}}\int \left\{1-\sin^2{\left(\theta\right)}\right\}^{-\frac{1}{2}}\;\cos{\left(\theta\right)}\mathrm{d}\theta \\&=&2\int \left\{1-\sin^2{\left(\theta\right)}\right\}^{-\frac{1}{2}}\;\cos{\left(\theta\right)}\mathrm{d}\theta \;\cdots\;\sqrt{A}A^{-\frac{1}{2}}=A^{\frac{1}{2}}A^{-\frac{1}{2}}=A^{-\frac{1}{2}+\frac{1}{2}}=A^0=1 \\&=&2\int \left\{\cos^2{\left(\theta\right)}\right\}^{-\frac{1}{2}}\;\cos{\left(\theta\right)}\mathrm{d}\theta \;\cdots\;\cos^2{\left(\theta\right)}+\sin^2{\left(\theta\right)}=1,\cos^2{\left(\theta\right)}=1-\sin^2{\left(\theta\right)} \\&=&2\int \left\{\cos{\left(\theta\right)}\right\}^{2(-\frac{1}{2})}\;\cos{\left(\theta\right)}\mathrm{d}\theta \;\cdots\;\left(A^B\right)^C=A^{BC} \\&=&2\int \left\{\cos{\left(\theta\right)}\right\}^{-1}\;\cos{\left(\theta\right)}\mathrm{d}\theta \\&=&2\int \mathrm{d}\theta \;\cdots\;A^{-1}A=1 \\&=&2\theta \;\cdots\;\int \mathrm{d}x=x+C\;(C:積分定数) \\&=&2\sin^{-1}{\left(\frac{u}{\sqrt{a}}\right)} \;\cdots\;\theta=\sin^{-1}{\left(\frac{u}{\sqrt{a}}\right)} \\&=&2\sin^{-1}{\left(\frac{\sqrt{x}}{\sqrt{a}}\right)} \;\cdots\;u=\sqrt{x} \\&=&2\sin^{-1}{\left(\sqrt{\frac{x}{a}}\right)}+C\;\cdots\;C:積分定数 \end{eqnarray} $$定積分
$$ \begin{eqnarray} \int_0^a \frac{1}{\sqrt{x(a-x)}}\mathrm{d}x &=&\left[2\sin^{-1}{\left(\sqrt{\frac{x}{a}}\right)}\right]_0^a \\&=&\left\{2\sin^{-1}{\left(\sqrt{\frac{a}{a}}\right)}\right\}-\left\{2\sin^{-1}{\left(\sqrt{\frac{0}{a}}\right)}\right\} \\&=&\left\{2\sin^{-1}{\left(1\right)}\right\}-\left\{2\sin^{-1}{\left(0\right)}\right\} \\&=&\left(2\frac{\pi}{2}\right)-\left(2\cdot0\right) \\&=&\pi-0 \\&=&\pi \end{eqnarray} $$sinの二乗の不定積分
sinの二乗の不定積分
$$ \begin{eqnarray} \int \sin^2{\left(\theta\right)}\mathrm{d}\theta &=&\int \frac{1}{2}-\frac{1}{2}\cos{\left(2\theta\right)}\mathrm{d}\theta \\&&\;\cdots\;\cos{\left(2\theta\right)}=\cos^2{\left(\theta\right)}-\sin^2{\left(\theta\right)}=\left(1-\sin^2{\left(\theta\right)}\right)-\sin^2{\left(\theta\right)}=1-2\sin^2{\left(\theta\right)} \\&&\;\cdots\;\sin^2{\left(\theta\right)}=\frac{1}{2}-\frac{1}{2}\cos{\left(2\theta\right)} \\&=&\frac{1}{2}\int \mathrm{d}\theta-\frac{1}{2}\int\cos{\left(2\theta\right)}\mathrm{d}\theta \\&=&\frac{1}{2}\theta-\frac{1}{2}\int\cos{\left(2\theta\right)}\mathrm{d}\theta \\&=&\frac{1}{2}\theta-\frac{1}{2}\int\cos{\left(\phi\right)}\frac{1}{2}\mathrm{d}\phi \;\cdots\;\phi=2\theta, \frac{\mathrm{d}\phi}{\mathrm{d}\theta}=2,\mathrm{d}\theta=\frac{1}{2}\mathrm{d}\phi \\&=&\frac{1}{2}\theta-\frac{1}{4}\int\cos{\left(\phi\right)}\mathrm{d}\phi\;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&\frac{1}{2}\theta-\frac{1}{2}\left(\frac{1}{2}\sin{\left(\phi\right)}\right) \;\cdots\;\int \cos{\left(x\right)}\mathrm{d}x= \sin{\left(x\right)}+C\;(C:積分定数) \\&=&\frac{1}{2}\theta-\frac{1}{4}\sin{\left(\phi\right)} \\&=&\frac{1}{2}\theta-\frac{1}{4}\sin{\left(2\theta\right)} \\&=&\frac{1}{2}\theta-\frac{1}{2}\sin{\left(\theta\right)}\cos{\left(\theta\right)} \\&=&\frac{1}{2}\left\{\theta-\sin{\left(\theta\right)}\cos{\left(\theta\right)}\right\}+C\;\cdots\;C:積分定数 \end{eqnarray} $$多角形D上の点から任意の点への距離の二乗の平均値(グリーンの定理を用いて面積分を周回積分にして求める)
グリーンの定理(Green's theorem)
閉曲線\(C\)で囲まれた領域\(D\)を考える場合，\(C^1\)級凾数\(P(x, y), Q(x, y)\)について以下が成り立つ。 $$ \begin{eqnarray} \iint_D \left(\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}\right)\mathrm{d}x\mathrm{d}y &=& \oint_C P(x, y)\mathrm{d}x+Q(x,y)\mathrm{d}y \end{eqnarray} $$面積分を周回積分へ変形し，区間毎の線積分に展開する
点\(x,y\)から点\((p,q)\)までの距離の二乗は $$ \begin{eqnarray} \langle r^2 \rangle&=&\left\{\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right\}^2 \\&=&\left(x-p\right)^2+\left(y-q\right)^2 \end{eqnarray} $$ であり，多角形D上の点(\(x,y\))から任意の点(\(p,q\))への距離の二乗の平均値\(\langle r^2 \rangle\)は， $$ \begin{eqnarray} \langle r^2 \rangle&=&\iint_D \rho(x,y)\left\{\left(x-p\right)^2+\left(y-q\right)^2\right\};\mathrm{d}x\mathrm{d}y \\&=&\iint_D \frac{1}{A}\left(x-p\right)^2+\left(y-q\right)^2\;\mathrm{d}x\mathrm{d}y \;\cdots\;密度は均一と仮定: \rho(x,y)=\frac{1}{A}\;(定数, A:Dの面積) \\&=&\frac{1}{A}\iint_D \left(x-p\right)^2+\left(y-q\right)^2\;\mathrm{d}x\mathrm{d}y \;\cdots\;\int_XcA(x)\mathrm{d}x=c\int_XA(x)\mathrm{d}x \end{eqnarray} $$ である．
これをグリーンの定理の式で満たすために例えば $$ \begin{eqnarray} Q(x, y)&=&\int \left(x-p\right)^2\;\mathrm{d}x \\P(x, y)&=&-\int \left(y-q\right)^2\;\mathrm{d}y \end{eqnarray} $$ とおく．これは $$ \begin{eqnarray} \frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y} &=& \int \left(x-p\right)^2\;\mathrm{d}x - \left\{-\int \left(y-q\right)^2\;\mathrm{d}y\right\} \\&=&\left(x-p\right)^2+\left(y-q\right)^2 \end{eqnarray} $$ となり，グリーンの定理の面積分側の被積分凾数を表現できている．
この\(P,Q\)を用いて線積分側の 被積分凾数を求めると以下のようになる． $$ \begin{eqnarray} \langle r^2 \rangle&=&\int_D \rho\left(x, y\right)r\left(x,y;p, q\right)^2 \;\mathrm{d}x\mathrm{d}y \\&=&\int_D \rho\left(x, y\right)\left\{\left(x-p\right)^2+\left(y-q\right)^2\right\} \;\mathrm{d}x\mathrm{d}y \\&=&\int_D \frac{1}{A}\left\{\left(x-p\right)^2+\left(y-q\right)^2\right\} \;\mathrm{d}x\mathrm{d}y \\&=&\frac{1}{A}\int_D \left\{\left(x-p\right)^2+\left(y-q\right)^2\right\} \;\mathrm{d}x\mathrm{d}y \\&=&\frac{1}{A}\left[ \int_D \left[ \frac{\partial }{\partial x} \left\{ \int \left(x-p\right)^2\;\mathrm{d}x \right\} - \frac{\partial }{\partial y} \left\{ -\int \left(y-q\right)^2\;\mathrm{d}y \right\} \right] \mathrm{d}x\mathrm{d}y \right] \\&=&\frac{1}{A}\left[ \oint_C \left\{ \int \left(x-p\right)^2\;\mathrm{d}x \right\} \mathrm{d}y + \left\{ -\int \left(y-q\right)^2\;\mathrm{d}y \right\} \mathrm{d}x \right] \\&=&\frac{1}{A}\left\{ \oint_C \frac{1}{3}\left(x-p\right)^3\;\mathrm{d}y - \frac{1}{3}\left(y-q\right)^3\;\mathrm{d}x \right\} \\&=&\frac{1}{A}\left\{ \oint_C \frac{1}{3}\left(x-p\right)^3\;\mathrm{d}y -\oint_C \frac{1}{3}\left(y-q\right)^3\;\mathrm{d}x \right\} \\&=&\frac{1}{A}\left\{ \frac{1}{3}\oint_C \left(x-p\right)^3\;\mathrm{d}y -\frac{1}{3}\oint_C \left(y-q\right)^3\;\mathrm{d}x \right\} \end{eqnarray} $$
\(\left\{\right\}\)内の第一項 区間毎の線積分
$$ \begin{eqnarray} \frac{1}{3}\oint_C \left(x-p\right)^3\;\mathrm{d}y &=& \frac{1}{3}\sum_{k=1}^n \int_{y_{k-1}}^{y_k} \left\{\left(\epsilon_k y+\zeta_k\right)-p\right\}^3\;\mathrm{d}y \;\cdots\;x=\epsilon_k y+\zeta_k 多角形の区間(頂点k-1からk)毎の直線の式 \\&=&\frac{1}{3}\sum_{k=1}^n \int_{y_{k-1}}^{y_k} \left\{ \left(\epsilon_k y+\zeta_k\right)^3 -3\left(\epsilon_k y+\zeta_k\right)^2 p +3\left(\epsilon_k y+\zeta_k\right) p^2 -p^3 \right\}\;\mathrm{d}y \;\cdots\;(A-B)^3=A^3-3A^2B+3AB^2-B^3 \\&=&\frac{1}{3}\sum_{k=1}^n \left[ \frac{1}{4\epsilon_k}\left(\epsilon_k y+\zeta_k\right)^4 -3\frac{1}{3\epsilon_k}\left(\epsilon_k y+\zeta_k\right)^3 p +3\frac{1}{2\epsilon_k}\left(\epsilon_k y+\zeta_k\right)^2 p^2 -y p^3 \right]_{y_{k-1}}^{y_k} \;\cdots\;\int cf(x) \mathrm{d}x= c\int f(x) \mathrm{d}x ,\;\int_a^b x^a \mathrm{d}x= \left[\frac{1}{a+1}x^{a+1}\right]_a^b \\&=&\frac{1}{3}\sum_{k=1} \left[ \frac{1}{4\epsilon_k}\left\{ \left(\epsilon_k y_k+\zeta_k\right)^4 -\left(\epsilon_k y_{k-1}+\zeta_k\right)^4 \right\} -\frac{1}{\epsilon_k}\left\{ \left(\epsilon_k y_k+\zeta_k\right)^3 -\left(\epsilon_k y_{k-1}+\zeta_k\right)^3 \right\} p +\frac{3}{2\epsilon_k}\left\{ \left(\epsilon_k y_{k}+\zeta_k\right)^2 -\left(\epsilon_k y_{k-1}+\zeta_k\right)^2 \right\} p^2 -\left( y_k - y_{k-1} \right) p^3 \right] \\&=&\frac{1}{3}\sum_{k=1}^n \left[ \frac{1}{4\epsilon_k}\left\{ \epsilon_k^4 (y_{k}^4-y_{k-1}^4) + 4 \epsilon_k^3 \zeta_k (y_{k}^3-y_{k-1}^3) + 6 \epsilon_k^2 \zeta_k^2 (y_{k}^2-y_{k-1}^2) + 4 \epsilon_k \zeta_k^3 (y_{k} -y_{k-1} ) \right\} -\frac{1}{\epsilon_k}\left\{ \epsilon_k^3 (y_{k}^3-y_{k-1}^3) + 3 \epsilon_k^2 \zeta_k (y_{k}^2-y_{k-1}^2) + 3 \epsilon_k \zeta_k^2 (y_{k} -y_{k-1} ) \right\} p +\frac{3}{2\epsilon_k}\left\{ \epsilon_k^2 (y_{k}^2-y_{k-1}^2) + 2 \epsilon_k \zeta_k (y_{k} -y_{k-1} ) \right\} p^2 -\left( y_k -y_{k-1} \right) p^3 \right] \\&&\;\cdots\;(a x + b)^4-(a y + b)^4 =a^4 x^4 - a^4 y^4 + 4 a^3 b x^3 - 4 a^3 b y^3 + 6 a^2 b^2 x^2 - 6 a^2 b^2 y^2 + 4 a b^3 x - 4 a b^3 y =a^4 (x^4 - y^4) + 4 a^3 b (x^3 - y^3) + 6 a^2 b^2 (x^2 - y^2) + 4 a b^3 (x - y) \\&&\;\cdots\;(a x + b)^3-(a y + b)^3 =a^3 x^3 - a^3 y^3 + 3 a^2 b x^2 - 3 a^2 b y^2 + 3 a b^2 x - 3 a b^2 y =a^3 (x^3 - y^3) + 3 a^2 b (x^2 - y^2) + 3 a b^2 (x - y) \\&&\;\cdots\;(a x + b)^2-(a y + b)^2 =a^2 x^2 - a^2 y^2 + 2 a b x - 2 a b y =a^2 (x^2 - y^2) + 2 a b (x - y) \\&=&\frac{1}{3}\sum_{k=1}^n \left[ \frac{1}{4\epsilon_k}\left\{ \epsilon_k^4 (y_{k}-y_{k-1})(y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 \epsilon_k^3 \zeta_k (y_{k}-y_{k-1})(y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 \epsilon_k^2 \zeta_k^2 (y_{k}-y_{k-1})(y_{k} +y_{k-1}) + 4 \epsilon_k \zeta_k^3 (y_{k}-y_{k-1}) \right\} -\frac{p}{\epsilon_k}\left\{ \epsilon_k^3 (y_{k}-y_{k-1})(y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 \epsilon_k^2 \zeta_k (y_{k}-y_{k-1})(y_{k} +y_{k-1}) + 3 \epsilon_k \zeta_k^2 (y_{k}-y_{k-1}) \right\} +\frac{3p^2}{2\epsilon_k}\left\{ \epsilon_k^2 (y_{k}-y_{k-1})(y_{k} +y_{k-1}) + 2 \epsilon_k \zeta_k (y_{k}-y_{k-1}) \right\} -p^3\left( y_k -y_{k-1} \right) \right] \\&&\;\cdots\;(a^4-b^4)=(a-b)(a+b)(a^2+b^2) \\&&\;\cdots\;(a^3-b^3)=(a-b)(a^2+ab+b^2) \\&&\;\cdots\;(a^2-b^2)=(a-b)(a+b) \\&=&\frac{1}{3}\sum_{k=1}^n (y_{k}-y_{k-1})\left[ \frac{1}{4\epsilon_k}\left\{ \epsilon_k^4 (y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 \epsilon_k^3 \zeta_k (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 \epsilon_k^2 \zeta_k^2 (y_{k} +y_{k-1}) + 4 \epsilon_k \zeta_k^3 \right\} -\frac{p}{\epsilon_k}\left\{ \epsilon_k^3 (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 \epsilon_k^2 \zeta_k (y_{k} +y_{k-1}) + 3 \epsilon_k \zeta_k^2 \right\} +\frac{3p^2}{2\epsilon_k}\left\{ \epsilon_k^2 (y_{k} +y_{k-1}) + 2 \epsilon_k \zeta_k \right\} -p^3 \right] \\&=&\frac{1}{3}\sum_{k=1}^n (y_{k}-y_{k-1})\left[ \frac{1}{4}\left\{ \epsilon_k^3 (y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 \epsilon_k^2 \zeta_k (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 \epsilon_k \zeta_k^2 (y_{k} +y_{k-1}) + 4 \zeta_k^3 \right\} -p\left\{ \epsilon_k^2 (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 \epsilon_k \zeta_k (y_{k} +y_{k-1}) + 3 \zeta_k^2 \right\} +\frac{3p^2}{2}\left\{ \epsilon_k (y_{k} +y_{k-1}) + 2 \zeta_k \right\} -p^3 \right] \end{eqnarray} $$\(\epsilon_k,\zeta_k\)の代入と整理
$$ \begin{eqnarray} \epsilon_k&=&\href{https://shikitenkai.blogspot.com/2020/07/blog-post_24.html}{\frac{x_{k}-x_{k-1}}{y_{k}-y_{k-1}}} \\ \zeta_k&=&\href{https://shikitenkai.blogspot.com/2020/07/blog-post_24.html}{\frac{(x_{k-1}y_{k}-x_{k}y_{k-1}) }{(y_{k}-y_{k-1})}} \\ \\&& \epsilon_k^3 (y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 \epsilon_k^2 \zeta_k (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 \epsilon_k \zeta_k^2 (y_{k} +y_{k-1}) + 4 \zeta_k^3 \\&=& \frac{(x_{k}-x_{k-1})^3}{(y_{k}-y_{k-1})^3} (y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 \frac{(x_{k}-x_{k-1})^2}{(y_{k}-y_{k-1})^2} \frac{(x_{k-1}y_{k}-x_{k}y_{k-1}) }{(y_{k}-y_{k-1}) } (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 \frac{(x_{k}-x_{k-1})}{(y_{k}-y_{k-1})} \frac{(x_{k-1}y_{k}-x_{k}y_{k-1})^2}{(y_{k}-y_{k-1})^2} (y_{k} +y_{k-1}) + 4 \frac{(x_{k-1}y_{k}-x_{k}y_{k-1})^3}{(y_{k}-y_{k-1})^3} \\&=&\frac{1}{(y_{k}-y_{k-1})^3} \left\{ (x_{k}-x_{k-1})^3 (y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 (x_{k}-x_{k-1})^2 (x_{k-1}y_{k}-x_{k}y_{k-1}) (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 (x_{k}-x_{k-1}) (x_{k-1}y_{k}-x_{k}y_{k-1})^2 (y_{k} +y_{k-1}) + 4 (x_{k-1}y_{k}-x_{k}y_{k-1})^3 \right\} \\&=& (x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2) \\ \\&& \epsilon_k^2 (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 \epsilon_k \zeta_k (y_{k} +y_{k-1}) + 3 \zeta_k^2 \\&=& \frac{(x_{k}-x_{k-1})^2}{(y_{k}-y_{k-1})^2} (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 \frac{(x_{k}-x_{k-1}) }{(y_{k}-y_{k-1}) } \frac{(x_{k-1}y_{k}-x_{k}y_{k-1}) }{(y_{k}-y_{k-1}) } (y_{k} +y_{k-1}) + 3 \frac{(x_{k-1}y_{k}-x_{k}y_{k-1})^2}{(y_{k}-y_{k-1})^2} \\&=&\frac{1}{(y_{k}-y_{k-1})^2} \left\{ (x_{k}-x_{k-1})^2 (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 (x_{k}-x_{k-1}) (x_{k-1}y_{k}-x_{k}y_{k-1}) (y_{k} +y_{k-1}) + 3 (x_{k-1}y_{k}-x_{k}y_{k-1})^2 \right\} \\&=&x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2 \\ \\&& \epsilon_k (y_{k} +y_{k-1}) + 2 \zeta_k \\&=& \frac{(x_{k}-x_{k-1}) }{(y_{k}-y_{k-1}) } (y_{k} +y_{k-1}) + 2 \frac{(x_{k-1}y_{k}-x_{k}y_{k-1}) }{(y_{k}-y_{k-1}) } \\&=&\frac{1}{(y_{k}-y_{k-1})} \left\{ (x_{k}-x_{k-1}) (y_{k} +y_{k-1}) + 2 (x_{k-1}y_{k}-x_{k}y_{k-1}) \right\} \\&=&x_{k}+x_{k-1} \end{eqnarray} $$\(\left\{\right\}\)内の第一項 整理後
$$ \begin{eqnarray} \frac{1}{3}\oint_C \left(x-p\right)^3\;\mathrm{d}y &=&\frac{1}{3}\sum_{k=1}^n (y_{k}-y_{k-1})\left[ \frac{1}{4}\left\{ \epsilon_k^3 (y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 \epsilon_k^2 \zeta_k (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 \epsilon_k \zeta_k^2 (y_{k} +y_{k-1}) + 4 \zeta_k^3 \right\} -p\left\{ \epsilon_k^2 (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 \epsilon_k \zeta_k (y_{k} +y_{k-1}) + 3 \zeta_k^2 \right\} +\frac{3p^2}{2}\left\{ \epsilon_k (y_{k} +y_{k-1}) + 2 \zeta_k \right\} -p^3 \right] \\&=&\frac{1}{3}\sum_{k=1}^n (y_{k}-y_{k-1})\left[ \frac{1}{4}\left\{(x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2)\right\} -p\left(x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2\right) +\frac{3p^2}{2}\left(x_{k}+x_{k-1}\right) -p^3 \right] \\&=& \frac{1}{12} \sum_{k=1}^n (y_{k}-y_{k-1}) \left\{(x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2)\right\} -\frac{p}{3} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2\right) +\frac{p^2}{2} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}+x_{k-1}\right) -\frac{p^3}{3} \sum_{k=1}^n (y_{k}-y_{k-1}) \\&=& \frac{1}{12} \sum_{k=1}^n (y_{k}-y_{k-1}) \left\{(x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2)\right\} -\frac{p}{3} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2\right) +\frac{p^2}{2} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}+x_{k-1}\right) \;\cdots\;\sum_{k=1}^n (y_{k}-y_{k-1}) = 0\;(周回積分なのでy軸の移動量の和は一周したら元に戻る) \end{eqnarray} $$\(\left\{\right\}\)内の第二項 整理後
$$ \begin{eqnarray} -\frac{1}{3}\oint_C \left(y-q\right)^3\;\mathrm{d}x &=& -\left[ \frac{1}{12} \sum_{k=1}^n (x_{k}-x_{k-1}) \left\{(y_{k}+y_{k-1})(y_{k}^2+y_{k-1}^2)\right\} -\frac{q}{3} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2\right) +\frac{q^2}{2} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}+y_{k-1}\right) \right] \end{eqnarray} $$項毎に計算したものをまとめる
$$ \begin{eqnarray} \langle r^2 \rangle &=& \frac{1}{A}\left\{ \frac{1}{3}\oint_C \left(x-p\right)^3\;\mathrm{d}y -\frac{1}{3}\oint_C \left(y-q\right)^3\;\mathrm{d}x \right\} \\&=& \frac{1}{A}\left[ \left[ \frac{1}{12} \sum_{k=1}^n (y_{k}-y_{k-1}) \left\{(x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2)\right\} -\frac{p}{3} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2\right) +\frac{p^2}{2} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}+x_{k-1}\right) \right] -\left[ \frac{1}{12} \sum_{k=1}^n (x_{k}-x_{k-1}) \left\{(y_{k}+y_{k-1})(y_{k}^2+y_{k-1}^2)\right\} -\frac{q}{3} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2\right) +\frac{q^2}{2} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}+y_{k-1}\right) \right] \right] \\&=& \frac{1}{A}\left[ \left[ \frac{1}{12} \sum_{k=1}^n (y_{k}-y_{k-1}) \left\{(x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2)\right\} -\frac{1}{12}\sum_{k=1}^n (x_{k}-x_{k-1}) \left\{(y_{k}+y_{k-1})(y_{k}^2+y_{k-1}^2)\right\} -2p \frac{1}{6} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2\right) -2q \left\{-\frac{1}{6} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2\right) \right\} +p^2\frac{1}{2} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}+x_{k-1}\right) +q^2\left\{-\frac{1}{2} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}+y_{k-1}\right)\right\} \right] \right] \\&=& \frac{1}{A}\left[ \lambda+\mu -2p A X_G-2q A Y_G +p^2 A+q^2 A \right] \\&&\;\cdots\;\lambda=\frac{1}{12} \sum_{k=1}^n (y_{k}-y_{k-1}) \left\{(x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2)\right\} \\&&\;\cdots\;\mu=-\frac{1}{12}\sum_{k=1}^n (x_{k}-x_{k-1}) \left\{(y_{k}+y_{k-1})(y_{k}^2+y_{k-1}^2)\right\} \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/blog-post_24.html}{X_G=\frac{1}{6A} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2\right)\;(重心X軸位置)} \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/blog-post_24.html}{Y_G=-\frac{1}{6A} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2\right)\;(重心Y軸位置)} \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/blog-post.html}{A= \frac{1}{2} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}+x_{k-1}\right)\;(面積)} \\&&\;\cdots\;A= -\frac{1}{2} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}+y_{k-1}\right)\;(面積) \\&=& \frac{\lambda+\mu}{A} -2p X_G -2q Y_G +p^2 +q^2 \\&=& \frac{\lambda+\mu}{A} {\color{red}+X_G^2} -2p X_G +p^2 {\color{red}+Y_G^2} -2q Y_G +q^2 {\color{red}-X_G^2} {\color{red}-Y_G^2} \\&=& \frac{\lambda+\mu}{A} +\left(p-X_G\right)^2 +\left(q-Y_G\right)^2 -X_G^2 -Y_G^2 \end{eqnarray} $$ よって\(\langle r^2 \rangle\)は多角形によって事前に決定される\(A,X_G,Y_G,\lambda,\mu\)と，点\((p,q)\)が与えられることで決まる．\(\langle r^2 \rangle\)の最小値
\(\langle r^2 \rangle\)の最小値となる点\((u, v)\)は，\(\left(p-X_G\right),\left(p-X_G\right)\)が0となる点\((X_G, Y_G)\)である．
また，点\((X_G, Y_G)\)から\(\langle r^2 \rangle\)の等しい点\((p, q)\)は，以下の式を満たすことになる． $$ \begin{eqnarray} \langle r^2 \rangle&=&\frac{\lambda+\mu}{A}+\left(p-X_G\right)^2+\left(q-Y_G\right)^2-X_G^2-Y_G^2 \\\left(p-X_G\right)^2+\left(q-Y_G\right)^2&=&\langle r^2 \rangle+X_G^2+Y_G^2-\frac{\lambda+\mu}{A} \\\sqrt{\left(p-X_G\right)^2+\left(q-Y_G\right)^2}&=&\sqrt{\langle r^2 \rangle+X_G^2+Y_G^2-\frac{\lambda+\mu}{A}} \end{eqnarray} $$
多角形D上の点から任意の点への距離の平均値 (グリーンの定理を用いて面積分を周回積分にして求める)
グリーンの定理(Green's theorem)
閉曲線\(C\)で囲まれた領域\(D\)を考える場合，\(C^1\)級凾数\(P(x, y), Q(x, y)\)について以下が成り立つ。 $$ \begin{eqnarray} \iint_D \left(\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}\right)\mathrm{d}x\mathrm{d}y &=& \oint_C P(x, y)\mathrm{d}x+Q(x,y)\mathrm{d}y \end{eqnarray} $$面積分を周回積分へ変形し，区間毎の積分に展開する
点\((x,y)\)から点\((p,q)\)までの距離は $$ \begin{eqnarray} r(x, y; p, q)&=&\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} \end{eqnarray} $$ であり，多角形D上の点\(x,y\)から任意の点\(p,q\)への距離の平均値\(\bar{r}\)は， $$ \begin{eqnarray} \bar{r}&=&\iint_D \rho(x,y)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x\mathrm{d}y \\&=&\iint_D \frac{1}{A}\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x\mathrm{d}y \;\cdots\;密度は均一と仮定: \rho(x,y)=\frac{1}{A}\;(定数, A:Dの面積) \\&=&\frac{1}{A}\iint_D \sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x\mathrm{d}y \;\cdots\;\int_XcA(x)\mathrm{d}x=c\int_XA(x)\mathrm{d}x \end{eqnarray} $$ である．これをグリーンの定理の式で満たすために例えば $$ \begin{eqnarray} Q(x, y)&=&\int \sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x \\P(x, y)&=&0 \end{eqnarray} $$ とおく．これは $$ \begin{eqnarray} \frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}&=&\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}-0 \\&=&\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} \end{eqnarray} $$ となり，グリーンの定理の面積分側の被積分凾数を表現できている．
この\(P,Q\)を用いて周回積分側の 被積分凾数を求めると以下のようになる． $$ \begin{eqnarray} \bar{r}&=&\frac{1}{A}\iint_D \sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x\mathrm{d}y \\&=&\frac{1}{A}\iint_D \left(\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}\right)\mathrm{d}x\mathrm{d}y \\&=&\frac{1}{A}\oint_C P(x, y)\mathrm{d}x+Q(x,y)\mathrm{d}y \\&=&\frac{1}{A}\oint_C 0\mathrm{d}x+\left(\int_D \sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x\right)\mathrm{d}y \\&=&\frac{1}{A}\oint_C \left(\int \sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x\right)\mathrm{d}y \\&=&\frac{1}{A}\oint_C \left(\int \sqrt{u^2+v^2}\;\mathrm{d}u\right)\mathrm{d}y \;\cdots\;u=x-p, v=y-q \\&=&\frac{1}{A}\oint_C \frac{1}{2}\left\{u\sqrt{u^2+v^2}+v^2\ln{\left|u+\sqrt{u^2+v^2}\right|}\right\}\mathrm{d}y \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/x2a2.html}{\int \sqrt{x^2+a^2}\;\mathrm{d}x=\frac{1}{2}\left\{x\sqrt{x^2+a^2}+a^2\ln{\left|x+\sqrt{x^2+a^2}\right|}\right\} + 積分定数} \\&=&\frac{1}{2A}\oint_C \left\{\left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} +\left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|}\right\}\mathrm{d}y \end{eqnarray} $$ ここで周回積分を多角形として考える．多角形の各頂点の列を以下のように与えるとする． $$ \begin{eqnarray} (x_0, y_0), (x_1, y_1),(x_2, y_2), \cdots, (x_n, y_n)=(x_0, y_0) \end{eqnarray} $$ 周回積分を多角形の区間ごとに分割し，区間毎の積分の和として以下のようになる． $$ \begin{eqnarray} \bar{r}&=&\frac{1}{2A}\oint_C \left\{\left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} +\left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|}\right\}\mathrm{d}y \\&=&\frac{1}{2A} \sum_{k=1}^n \int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left\{\left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} +\left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|}\right\}\mathrm{d}y \\&=&\frac{1}{2A} \sum_{k=1}^n\left\{ \int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} \;\mathrm{d}y +\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|} \;\mathrm{d}y \right\} \\&=&\frac{1}{2A} \sum_{k=1}^n \left\{I_{1k}+ I_{2k}\right\} \\&&\;\cdots\;I_{1k}=\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} \;\mathrm{d}y \\&&\;\cdots\;I_{2k}=\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|} \;\mathrm{d}y \end{eqnarray} $$
多角形の頂点間線分(辺)に重なる直線の式を求める
多角形の各頂点において今の頂点\((x_k, y_k)\)とその前の頂点\((x_{k-1}, y_{k-1})\)とを通る直線の式は以下のように求められる． $$ \begin{eqnarray} y-y_{k-1}&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}\left(x-x_{k-1}\right) \;\cdots\;傾き\thetaの直線における原点を，点(p,q)へ移動した式と同じ．(y-q)=\theta(x-p) \\x-x_{k-1}&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right) \end{eqnarray} $$ $$ \begin{eqnarray} \\x-x_{k-1}&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right) \\x&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right)+x_{k-1} \\x&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_k-x_{k-1}}{y_k-y_{k-1}}(-y_{k-1})+x_{k-1} \\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{\left(x_k-x_{k-1}\right)(-y_{k-1})+x_{k-1}\left(y_k-y_{k-1}\right)}{y_k-y_{k-1}} \\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y +\frac{ -x_ky_{k-1} +x_{k-1}y_{k-1} +x_{k-1}y_k -x_{k-1}y_{k-1} }{y_k-y_{k-1}} \\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_ky_{k-1}+x_{k-1}y_k}{y_k-y_{k-1}} \\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_{k-1}y_k-x_ky_{k-1}}{y_k-y_{k-1}} \\&=&\epsilon_k y+\zeta_k \end{eqnarray} $$区間毎の積分より距離の平均値\(\bar{r}\)を求める 第一項\(I_{1k}\)
$$ \begin{eqnarray} I_{1k}&=&\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} \;\mathrm{d}y \\&=&I_{1k}(y_k)-I_{1k}(y_{k-1}) \\I_{1k}(y)&=&\int \left(\left(\epsilon_k y+\zeta_k\right)-p\right) \sqrt{\left(\left(\epsilon_k y+\zeta_k\right)-p\right)^2+\left(y-q\right)^2} \;\mathrm{d}y \;\cdots\;x=\epsilon_k y+\zeta_k \\&=&\int \left(\left(\epsilon_k y+\zeta_k\right)-p\right) \sqrt{\left(\left(\epsilon_k y+\zeta_k\right)^2-2p\left(\epsilon_k y+\zeta_k\right)+p^2\right)+\left(y^2-2qy+q^2\right)} \;\mathrm{d}y \\&=&\int \left(\epsilon_k y+\zeta_k-p\right) \sqrt{\left(\epsilon_k^2 y^2+2\epsilon_k\zeta_k y+\zeta_k^2\right)-2p\epsilon_k y-2p\zeta_k+p^2+y^2-2qy+q^2} \;\mathrm{d}y \\&=&\int \left(\epsilon_k y+\zeta_k-p\right) \sqrt{\epsilon_k^2 y^2+2\epsilon_k\zeta_k y+\zeta_k^2-2p\epsilon_k y-2p\zeta_k+p^2+y^2-2qy+q^2} \;\mathrm{d}y \\&=&\int \left(\epsilon_k y+\zeta_k-p\right) \sqrt{\left(\epsilon_k^2 +1 \right)y^2+2\left\{\epsilon_k\left(\zeta_k-p\right)-q\right\}y +\left(\zeta_k-p\right)^2+q^2} \;\mathrm{d}y \\&=&\int \left(\epsilon_k y+\zeta_k-p\right)\sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y \;\cdots\;\alpha_k=\epsilon_k^2 +1,\;\beta_k=2\left\{\epsilon_k\left(\zeta_k-p\right)-q\right\},\;\gamma_k=\left(\zeta_k-p\right)^2+q^2 \\&=& \int \epsilon_k y\sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y + \int \left(\zeta_k-p\right) \sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y \\&=& \epsilon_k \int y\sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y + \left(\zeta_k-p\right) \int \sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y \\&=& \epsilon_k \int y\sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y + \left(\zeta_k-p\right) \int \sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y \\&=& \epsilon_k \left\{ \frac{1}{3\alpha_k}\left(\alpha_k y^2+\beta_k y+\gamma_k \right)^{\frac{3}{2}} -\frac{\beta_k \left(2\alpha_k y+\beta_k \right)}{8\alpha_k^{2}}\sqrt{\alpha_k y^2+\beta_k y+\gamma_k} +\frac{\beta_k \left(\beta_k^2-4\alpha_k\gamma_k\right)}{16\alpha_k^{\frac{5}{2}}}\ln{ \left|\left(2\alpha_k y+\beta_k \right)+2\sqrt{\alpha_k\left(\alpha_k y^2+\beta_k y+\gamma_k\right)}\right| } +C_0 \right\} \\&&+ \left(\zeta_k-p\right) \left\{ \frac{2\alpha_k y+\beta_k}{4\alpha_k}\sqrt{\alpha_k y^2+\beta_k y+\gamma_k} +\frac{4\alpha_k \gamma_k-\beta_k^2}{8\alpha_k^{\frac{3}{2}}}\ln{ \left|\left(2\alpha_k y+\beta_k\right)+2\sqrt{\alpha_k\left(\alpha_k y^2+\beta_k y+\gamma_k\right)}\right| } +C_1 \right\} \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/xa-x2b-x-c.html}{\int x\sqrt{a x^2+b x +c} \;\mathrm{d}x= \frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}} -\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } +C_0\;(C_0:積分定数)} \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/a-x2b-x-c.html}{\int \sqrt{a x^2+b x +c} \;\mathrm{d}x= \frac{2ax+b}{4a}\sqrt{\left(ax^2+bx+c\right)} +\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } +C_1\;(C_1:積分定数)} \\&=& \epsilon_k \left\{ \frac{1}{3\alpha_k}\left(\alpha_k y^2+\beta_k y+\gamma_k \right)^{\frac{3}{2}} -\frac{\beta_k \left(2\alpha_k y+\beta_k \right)}{8\alpha_k^{2}}\sqrt{\alpha_k y^2+\beta_k y+\gamma_k} +\frac{\beta_k \left(\beta_k^2-4\alpha_k\gamma_k\right)}{16\alpha_k^{\frac{5}{2}}}\ln{ \left|\left(2\alpha_k y+\beta_k \right)+2\sqrt{\alpha_k\left(\alpha_k y^2+\beta_k y+\gamma_k\right)}\right| } \right\} \\&&+ \left(\zeta_k-p\right) \left\{ \frac{2\alpha_k y+\beta_k}{4\alpha_k}\sqrt{\alpha_k y^2+\beta_k y+\gamma_k} +\frac{4\alpha_k \gamma_k-\beta_k^2}{8\alpha_k^{\frac{3}{2}}}\ln{ \left|\left(2\alpha_k y+\beta_k\right)+2\sqrt{\alpha_k\left(\alpha_k y^2+\beta_k y+\gamma_k\right)}\right| } \right\} +\epsilon_k C_0 +\left(\zeta_k-p\right) C_1 \\&=& \epsilon_k \left\{ \frac{1}{3\alpha_k}\left(\alpha_k y^2+\beta_k y+\gamma_k \right)^{\frac{3}{2}} -\frac{\beta_k \left(2\alpha_k y+\beta_k \right)}{8\alpha_k^{2}}\sqrt{\alpha_k y^2+\beta_k y+\gamma_k} +\frac{\beta_k \left(\beta_k^2-4\alpha_k\gamma_k\right)}{16\alpha_k^{\frac{5}{2}}}\ln{ \left|\left(2\alpha_k y+\beta_k \right)+2\sqrt{\alpha_k\left(\alpha_k y^2+\beta_k y+\gamma_k\right)}\right| } \right\} \\&&+ \left(\zeta_k-p\right) \left\{ \frac{2\alpha_k y+\beta_k}{4\alpha_k}\sqrt{\alpha_k y^2+\beta_k y+\gamma_k} +\frac{4\alpha_k \gamma_k-\beta_k^2}{8\alpha_k^{\frac{3}{2}}}\ln{ \left|\left(2\alpha_k y+\beta_k\right)+2\sqrt{\alpha_k\left(\alpha_k y^2+\beta_k y+\gamma_k\right)}\right| } \right\} + C \;\cdots\;C=\epsilon_k C_0+\left(\zeta_k-p\right) C_1\;(C:積分定数) \end{eqnarray} $$区間毎の積分より距離の平均値\(\bar{r}\)を求める 第二項\(I_{2k}\)
$$ \begin{eqnarray} I_{2k}&=&\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|} \;\mathrm{d}y \\&=&\int^{y_k}_{y_{k-1}} \left(y-q\right)^2\ln{\left|\left(\left(\epsilon_k y+\zeta_k\right)-p\right)+\sqrt{\left(\left(\epsilon_k y+\zeta_k\right)-p\right)^2+\left(y-q\right)^2}\right|} \;\mathrm{d}y \\&=&\int^{y_k}_{y_{k-1}} \left(y-q\right)^2 \ln{\left| \epsilon_k y+\zeta_k-p +\sqrt{ \left( \left( \epsilon_k y+\zeta_k \right)^2 -2p\left( \epsilon_k y+\zeta_k \right)+p^2 \right) +\left( y^2-2qy-q^2 \right) } \right|} \;\mathrm{d}y \\&=&\int^{y_k}_{y_{k-1}} \left(y-q\right)^2 \ln{\left| \epsilon_k y+\zeta_k-p +\sqrt{ \left( \epsilon_k^2 y^2 +2\epsilon_k\zeta_k y +\zeta_k^2 \right) -2p\epsilon_k y -2p\zeta_k +p^2 +y^2-2qy-q^2 } \right|} \;\mathrm{d}y \\&=&\int^{y_k}_{y_{k-1}} \left(y-q\right)^2 \ln{\left| \epsilon_k y+\zeta_k-p +\sqrt{ \epsilon_k^2 y^2 +y^2 +2\epsilon_k\zeta_k y -2p\epsilon_k y -2qy +\zeta_k^2 -2p\zeta_k +p^2 -q^2 } \right|} \;\mathrm{d}y \\&=&\int^{y_k}_{y_{k-1}} \left(y-q\right)^2 \ln{\left| \epsilon_k y+\zeta_k-p +\sqrt{ \left(\epsilon_k^2+1\right) y^2 +2\left\{\epsilon_k\left(\zeta_k-p\right)-q\right\} y +\left(\zeta_k-p\right)^2 -q^2 } \right|} \;\mathrm{d}y \\&=&\int^{y_k}_{y_{k-1}} \left(y-q\right)^2 \ln{\left| \epsilon_k y+\zeta_k-p +\sqrt{\alpha_k y^2+\beta_k y+\gamma_k} \right|}\;\mathrm{d}y \;\cdots\;\alpha_k=\epsilon_k^2+1 ,\;\beta_k=2\left\{\epsilon_k \left(\zeta_k-p\right) -q\right\} ,\;\gamma_k=\left(\zeta_k-p\right)^2-q^2 \end{eqnarray} $$ →要:数値計算区間毎の積分より距離の平均値\(\bar{r}\)を求める 第一項\(I_{1k}\)+第二項\(I_{2k}\)
$$ \begin{eqnarray} \bar{r}&=&\frac{1}{2A} \sum_{k=1}^n\left\{ \int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} \;\mathrm{d}y +\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|} \;\mathrm{d}y \right\} \\&=&\frac{1}{2A} \sum_{k=1}^n \left\{I_{1k} + I_{2k}\right\} \\&=&\frac{1}{2A} \sum_{k=1}^n \left\{I_{1k}(y_{k}) - I_{1k}(y_{k-1}) + I_{2k}\right\} \end{eqnarray} $$x√(a x^2+b x +c)の積分
$$
\begin{eqnarray}
\int x\sqrt{a x^2+b x +c} \;\mathrm{d}x
&=&\int x\sqrt{\left(\sqrt{a} x+\frac{b}{2\sqrt{a}} \right)^2-\frac{b^2}{4a}+c} \;\mathrm{d}x
\;\cdots\;平方完成
\\&&\;\cdots\;ax^2+bx+c=(\sqrt{a}x+\alpha)^2-\alpha^2+c
\\&&\;\cdots\;(\sqrt{a}x+\alpha)^2=ax^2+2\sqrt{a}\alpha x+\alpha^2,\;2\sqrt{a}\alpha=bより\alpha=\frac{b}{2\sqrt{a}}
\\&=&\int \frac{2\sqrt{a}u-b}{2a}\; \sqrt{u^2-\frac{b^2}{4a}+c} \;\frac{1}{\sqrt{a}}\mathrm{d}u
\\&&\;\cdots\;u=\sqrt{a} x+\frac{b}{2\sqrt{a}}
\\&&\;\cdots\;x=\frac{u-\frac{b}{2\sqrt{a}}}{\sqrt{a}}
=\frac{\frac{2\sqrt{a}u-b}{2\sqrt{a}}}{\sqrt{a}}
=\frac{2\sqrt{a}u-b}{2a}
\\&&\;\cdots\;\frac{\mathrm{d}x}{\mathrm{d}u}=\frac{1}{\sqrt{a}},\mathrm{d}x=\frac{1}{\sqrt{a}}\mathrm{d}u
\\&=&\frac{1}{2a^{\frac{3}{2}}} \int \left(2\sqrt{a}u-b\right)\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u
\\&=& \frac{1}{2a^{\frac{3}{2}}}\int 2\sqrt{a}u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u
- \frac{1}{2a^{\frac{3}{2}}} \int b\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u
\\&=& \frac{2\sqrt{a}}{2a^{\frac{3}{2}}}\int u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u
- \frac{b}{2a^{\frac{3}{2}}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u
\\&=& \frac{1}{a}\int u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u
- \frac{b}{2a^{\frac{3}{2}}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u
\end{eqnarray}
$$
$$ \begin{eqnarray} \frac{1}{a}\int u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u &=& \frac{1}{a}\int u\;\sqrt{s} \;\frac{1}{2u}\;\mathrm{d}s \;\cdots\;s=u^2-\frac{b^2}{4a}+c,\frac{\mathrm{d}s}{\mathrm{d}u}=2u,\mathrm{d}u=\frac{1}{2u}\;\mathrm{d}s \\&=& \frac{1}{2a}\int\sqrt{s}\;\mathrm{d}s \\&=& \frac{1}{2a}\left[\frac{2}{3}s^{\frac{3}{2}}+C_0\right]\;\cdots\;\int x^{\frac{1}{2}}\;\mathrm{d}x=\frac{2}{3}x^{\frac{3}{2}}+C_0\;(C_0:積分定数) \\&=& \frac{1}{3a}s^{\frac{3}{2}}+\frac{1}{2a}C_0 \\&=& \frac{1}{3a}\left(u^2-\frac{b^2}{4a}+c\right)^{\frac{3}{2}}+\frac{1}{2a}C_0 \\&=& \frac{1}{3a}\left\{\left(\sqrt{a} x+\frac{b}{2\sqrt{a}}\right)^2-\frac{b^2}{4a}+c\right\}^{\frac{3}{2}}+\frac{1}{2a}C_0 \\&=& \frac{1}{3a}\left\{\left(ax^2+bx+\frac{b^2}{4a}\right)-\frac{b^2}{4a}+c\right\}^{\frac{3}{2}}+\frac{1}{2a}C_0 \\&=& \frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}}+\frac{1}{2a}C_0 \end{eqnarray} $$
$$ \begin{eqnarray} -\frac{b}{2a^{\frac{3}{2}}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u &=&-\frac{b}{2a^{\frac{3}{2}}} \int \sqrt{\left(c-\frac{b^2}{4a}\right)\left(t^2+1\right)} \;\sqrt{c-\frac{b^2}{4a}}\;\mathrm{d}t \\&&\;\cdots\;t=\frac{u}{\sqrt{c-\frac{b^2}{4a}}} ,\frac{\mathrm{d}t}{\mathrm{d}u}=\frac{1}{\sqrt{c-\frac{b^2}{4a}}} ,\mathrm{d}u=\sqrt{c-\frac{b^2}{4a}}\mathrm{d}t \\&&\;\cdots\;u^2+c-\frac{b^2}{4a} =\left(u^2+c-\frac{b^2}{4a}\right)\frac{c-\frac{b^2}{4a}}{c-\frac{b^2}{4a}} =\left(c-\frac{b^2}{4a}\right)\frac{u^2+c-\frac{b^2}{4a}}{c-\frac{b^2}{4a}} =\left(c-\frac{b^2}{4a}\right)\left(\frac{u^2}{c-\frac{b^2}{4a}}+1\right) =\left(c-\frac{b^2}{4a}\right)\left\{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1\right\} =\left(c-\frac{b^2}{4a}\right)\left(t^2+1\right) \\&=& - \frac{b}{2a^{\frac{3}{2}}} \left(c-\frac{b^2}{4a}\right)\int \sqrt{t^2+1} \;\mathrm{d}t \\&=& - \frac{b}{2a^{\frac{3}{2}}} \left(c-\frac{b^2}{4a}\right) \left\{\frac{1}{2}\left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_1\right)\right\} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/x2a2.html}{\int \sqrt{x^2+a^2} \;\mathrm{d}x = \frac{1}{2}\left(x\sqrt{x^2+a^2}+ a^2 \ln{ \left|x+\sqrt{x^2+a^2}\right| }+C_1\right)\;(C_1:積分定数)} \\&=& \left(\frac{b^3}{16a^{\frac{5}{2}}}-\frac{bc}{4a^{\frac{3}{2}}}\right) \left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_1\right) \\&=& \left(\frac{b^3}{16a^{\frac{5}{2}}}-\frac{4abc}{16a^{\frac{5}{2}}}\right) \left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_1\right) \\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}} \left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_1\right) \\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}} \left\{ \left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)\sqrt{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1} +\ln{ \left|\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)+\sqrt{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1}\right| }+C_1 \right\} \\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}} \left\{ \left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)\sqrt{\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1} +\ln{ \left|\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)+\sqrt{\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1}\right| }+C_1 \right\} \\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}} \left\{ \frac{2ax+b}{\sqrt{4ac-b^2}}\sqrt{\frac{4a\left(ax^2+bx+c\right)}{4ac-b^2}} +\ln{ \left|\frac{2ax+b}{\sqrt{4ac-b^2}}+\sqrt{\frac{4a\left(ax^2+bx+c\right)}{4ac-b^2}}\right| }+C_1 \right\} \\&&\;\cdots\;\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}=\frac{2ax+b}{\sqrt{4ac-b^2}} \\&&\;\cdots\;\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2=\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)^2=\frac{4a^2x^2+4abx+b^2}{4ac-b^2} \\&&\;\cdots\;\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1=\frac{4a^2x^2+4abx+b^2}{4ac-b^2}+1=\frac{4a^2x^2+4abx+b^2+4ac-b^2}{4ac-b^2}=\frac{4a^2x^2+4abx+4ac}{4ac-b^2}=\frac{4a\left(ax^2+bx+c\right)}{4ac-b^2} \\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}} \left\{ \frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{\left(ax^2+bx+c\right)} +\ln{ \left|\frac{1}{\sqrt{4ac-b^2}}\left\{\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right\}\right| } +C_1 \right\} \\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}} \left\{ \frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{a\left(ax^2+bx+c\right)} +\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\ln{\sqrt{4ac-b^2}} +C_1 \right\} \\&=&\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1 \\&=&-\frac{b\left(2ax+b\right)}{8a^{\frac{4}{2}}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1 \\&=&-\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1 \end{eqnarray} $$
$$ \begin{eqnarray} \int x\sqrt{a x^2+b x +c} \;\mathrm{d}x &=& \frac{1}{a}\int u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u - \frac{b}{2a^{\frac{3}{2}}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u \\&=& \frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}}+\frac{1}{2a}C_0 -\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1 \\&&\;\cdots\;\frac{1}{a}\int u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u=\frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}}+\frac{1}{2a}C_0 \\&&\;\cdots\;-\frac{b}{2a^{\frac{3}{2}}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u=-\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1 \\&=& \frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}} -\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } +\frac{1}{2a}C_0 -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1 \\&=& \frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}} -\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } +C \;\cdots\;C=\frac{1}{2a}C_0 -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1\;(C:積分定数) \end{eqnarray} $$
$$ \begin{eqnarray} \frac{1}{a}\int u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u &=& \frac{1}{a}\int u\;\sqrt{s} \;\frac{1}{2u}\;\mathrm{d}s \;\cdots\;s=u^2-\frac{b^2}{4a}+c,\frac{\mathrm{d}s}{\mathrm{d}u}=2u,\mathrm{d}u=\frac{1}{2u}\;\mathrm{d}s \\&=& \frac{1}{2a}\int\sqrt{s}\;\mathrm{d}s \\&=& \frac{1}{2a}\left[\frac{2}{3}s^{\frac{3}{2}}+C_0\right]\;\cdots\;\int x^{\frac{1}{2}}\;\mathrm{d}x=\frac{2}{3}x^{\frac{3}{2}}+C_0\;(C_0:積分定数) \\&=& \frac{1}{3a}s^{\frac{3}{2}}+\frac{1}{2a}C_0 \\&=& \frac{1}{3a}\left(u^2-\frac{b^2}{4a}+c\right)^{\frac{3}{2}}+\frac{1}{2a}C_0 \\&=& \frac{1}{3a}\left\{\left(\sqrt{a} x+\frac{b}{2\sqrt{a}}\right)^2-\frac{b^2}{4a}+c\right\}^{\frac{3}{2}}+\frac{1}{2a}C_0 \\&=& \frac{1}{3a}\left\{\left(ax^2+bx+\frac{b^2}{4a}\right)-\frac{b^2}{4a}+c\right\}^{\frac{3}{2}}+\frac{1}{2a}C_0 \\&=& \frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}}+\frac{1}{2a}C_0 \end{eqnarray} $$
$$ \begin{eqnarray} -\frac{b}{2a^{\frac{3}{2}}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u &=&-\frac{b}{2a^{\frac{3}{2}}} \int \sqrt{\left(c-\frac{b^2}{4a}\right)\left(t^2+1\right)} \;\sqrt{c-\frac{b^2}{4a}}\;\mathrm{d}t \\&&\;\cdots\;t=\frac{u}{\sqrt{c-\frac{b^2}{4a}}} ,\frac{\mathrm{d}t}{\mathrm{d}u}=\frac{1}{\sqrt{c-\frac{b^2}{4a}}} ,\mathrm{d}u=\sqrt{c-\frac{b^2}{4a}}\mathrm{d}t \\&&\;\cdots\;u^2+c-\frac{b^2}{4a} =\left(u^2+c-\frac{b^2}{4a}\right)\frac{c-\frac{b^2}{4a}}{c-\frac{b^2}{4a}} =\left(c-\frac{b^2}{4a}\right)\frac{u^2+c-\frac{b^2}{4a}}{c-\frac{b^2}{4a}} =\left(c-\frac{b^2}{4a}\right)\left(\frac{u^2}{c-\frac{b^2}{4a}}+1\right) =\left(c-\frac{b^2}{4a}\right)\left\{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1\right\} =\left(c-\frac{b^2}{4a}\right)\left(t^2+1\right) \\&=& - \frac{b}{2a^{\frac{3}{2}}} \left(c-\frac{b^2}{4a}\right)\int \sqrt{t^2+1} \;\mathrm{d}t \\&=& - \frac{b}{2a^{\frac{3}{2}}} \left(c-\frac{b^2}{4a}\right) \left\{\frac{1}{2}\left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_1\right)\right\} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/x2a2.html}{\int \sqrt{x^2+a^2} \;\mathrm{d}x = \frac{1}{2}\left(x\sqrt{x^2+a^2}+ a^2 \ln{ \left|x+\sqrt{x^2+a^2}\right| }+C_1\right)\;(C_1:積分定数)} \\&=& \left(\frac{b^3}{16a^{\frac{5}{2}}}-\frac{bc}{4a^{\frac{3}{2}}}\right) \left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_1\right) \\&=& \left(\frac{b^3}{16a^{\frac{5}{2}}}-\frac{4abc}{16a^{\frac{5}{2}}}\right) \left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_1\right) \\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}} \left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_1\right) \\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}} \left\{ \left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)\sqrt{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1} +\ln{ \left|\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)+\sqrt{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1}\right| }+C_1 \right\} \\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}} \left\{ \left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)\sqrt{\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1} +\ln{ \left|\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)+\sqrt{\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1}\right| }+C_1 \right\} \\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}} \left\{ \frac{2ax+b}{\sqrt{4ac-b^2}}\sqrt{\frac{4a\left(ax^2+bx+c\right)}{4ac-b^2}} +\ln{ \left|\frac{2ax+b}{\sqrt{4ac-b^2}}+\sqrt{\frac{4a\left(ax^2+bx+c\right)}{4ac-b^2}}\right| }+C_1 \right\} \\&&\;\cdots\;\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}=\frac{2ax+b}{\sqrt{4ac-b^2}} \\&&\;\cdots\;\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2=\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)^2=\frac{4a^2x^2+4abx+b^2}{4ac-b^2} \\&&\;\cdots\;\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1=\frac{4a^2x^2+4abx+b^2}{4ac-b^2}+1=\frac{4a^2x^2+4abx+b^2+4ac-b^2}{4ac-b^2}=\frac{4a^2x^2+4abx+4ac}{4ac-b^2}=\frac{4a\left(ax^2+bx+c\right)}{4ac-b^2} \\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}} \left\{ \frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{\left(ax^2+bx+c\right)} +\ln{ \left|\frac{1}{\sqrt{4ac-b^2}}\left\{\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right\}\right| } +C_1 \right\} \\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}} \left\{ \frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{a\left(ax^2+bx+c\right)} +\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\ln{\sqrt{4ac-b^2}} +C_1 \right\} \\&=&\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1 \\&=&-\frac{b\left(2ax+b\right)}{8a^{\frac{4}{2}}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1 \\&=&-\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1 \end{eqnarray} $$
$$ \begin{eqnarray} \int x\sqrt{a x^2+b x +c} \;\mathrm{d}x &=& \frac{1}{a}\int u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u - \frac{b}{2a^{\frac{3}{2}}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u \\&=& \frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}}+\frac{1}{2a}C_0 -\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1 \\&&\;\cdots\;\frac{1}{a}\int u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u=\frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}}+\frac{1}{2a}C_0 \\&&\;\cdots\;-\frac{b}{2a^{\frac{3}{2}}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u=-\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1 \\&=& \frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}} -\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } +\frac{1}{2a}C_0 -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1 \\&=& \frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}} -\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } +C \;\cdots\;C=\frac{1}{2a}C_0 -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1\;(C:積分定数) \end{eqnarray} $$
√(a x^2+b x +c)の積分
$$
\begin{eqnarray}
\int \sqrt{a x^2+b x +c} \;\mathrm{d}x
&=&\int \sqrt{\left(\sqrt{a} x+\frac{b}{2\sqrt{a}} \right)^2-\frac{b^2}{4a}+c} \;\mathrm{d}x
\;\cdots\;平方完成
\\&&\;\cdots\;ax^2+bx+c=(\sqrt{a}x+\alpha)^2-\alpha^2+c
\\&&\;\cdots\;(\sqrt{a}x+\alpha)^2=ax^2+2\sqrt{a}\alpha x+\alpha^2,\;2\sqrt{a}\alpha=bより\alpha=\frac{b}{2\sqrt{a}}
\\&=&\int \sqrt{u^2-\frac{b^2}{4a}+c} \;\frac{1}{\sqrt{a}}\mathrm{d}u
\\&&\;\cdots\;u=\sqrt{a} x+\frac{b}{2\sqrt{a}}
\\&&\;\cdots\;\frac{\mathrm{d}u}{\mathrm{d}x}=\sqrt{a},\mathrm{d}x=\frac{1}{\sqrt{a}}\mathrm{d}u
\\&=& \frac{1}{\sqrt{a}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u
\\&=& \frac{1}{\sqrt{a}} \int \sqrt{\left(c-\frac{b^2}{4a}\right)\left(t^2+1\right)} \;\sqrt{c-\frac{b^2}{4a}}\;\mathrm{d}t
\\&&\;\cdots\;t=\frac{u}{\sqrt{c-\frac{b^2}{4a}}}
,\frac{\mathrm{d}t}{\mathrm{d}u}=\frac{1}{\sqrt{c-\frac{b^2}{4a}}}
,\mathrm{d}u=\sqrt{c-\frac{b^2}{4a}}\mathrm{d}t
\\&&\;\cdots\;u^2+c-\frac{b^2}{4a}
=\left(u^2+c-\frac{b^2}{4a}\right)\frac{c-\frac{b^2}{4a}}{c-\frac{b^2}{4a}}
=\left(c-\frac{b^2}{4a}\right)\frac{u^2+c-\frac{b^2}{4a}}{c-\frac{b^2}{4a}}
=\left(c-\frac{b^2}{4a}\right)\left(\frac{u^2}{c-\frac{b^2}{4a}}+1\right)
=\left(c-\frac{b^2}{4a}\right)\left\{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1\right\}
=\left(c-\frac{b^2}{4a}\right)\left(t^2+1\right)
\\&=& \frac{4ac-b^2}{4a^{\frac{3}{2}}} \int \sqrt{t^2+1} \;\mathrm{d}t
\\&=& \frac{4ac-b^2}{4a^{\frac{3}{2}}}
\left\{\frac{1}{2}\left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_0\right)\right\}
\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/x2a2.html}{\int \sqrt{x^2+a^2} \;\mathrm{d}x = \frac{1}{2}\left(x\sqrt{x^2+a^2}+ a^2 \ln{ \left|x+\sqrt{x^2+a^2}\right| }+C_0\right)\;(C_0:積分定数)}
\\&=& \frac{4ac-b^2}{8a^{\frac{3}{2}}}
\left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_0\right)
\\&=& \frac{4ac-b^2}{8a^{\frac{3}{2}}}
\left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_0\right)
\\&=& \frac{4ac-b^2}{8a^{\frac{3}{2}}}
\left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_0\right)
\\&=& \frac{4ac-b^2}{8a^{\frac{3}{2}}}
\left\{
\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)\sqrt{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1}
+\ln{ \left|\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)+\sqrt{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1}\right| }+C_0
\right\}
\\&=& \frac{4ac-b^2}{8a^{\frac{3}{2}}}
\left\{
\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)\sqrt{\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1}
+\ln{ \left|\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)+\sqrt{\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1}\right| }+C_0
\right\}
\\&=& \frac{4ac-b^2}{8a^{\frac{3}{2}}}
\left\{
\frac{2ax+b}{\sqrt{4ac-b^2}}\sqrt{\frac{4a\left(ax^2+bx+c\right)}{4ac-b^2}}
+\ln{ \left|\frac{2ax+b}{\sqrt{4ac-b^2}}+\sqrt{\frac{4a\left(ax^2+bx+c\right)}{4ac-b^2}}\right| }+C_0
\right\}
\\&&\;\cdots\;\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}=\frac{2ax+b}{\sqrt{4ac-b^2}}
\\&&\;\cdots\;\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2=\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)^2=\frac{4a^2x^2+4abx+b^2}{4ac-b^2}
\\&&\;\cdots\;\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1=\frac{4a^2x^2+4abx+b^2}{4ac-b^2}+1=\frac{4a^2x^2+4abx+b^2+4ac-b^2}{4ac-b^2}=\frac{4a^2x^2+4abx+4ac}{4ac-b^2}=\frac{4a\left(ax^2+bx+c\right)}{4ac-b^2}
\\&=& \frac{4ac-b^2}{8a^{\frac{3}{2}}}
\left\{
\frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{\left(ax^2+bx+c\right)}
+\ln{ \left|\frac{1}{\sqrt{4ac-b^2}}\left\{\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right\}\right| }
+C_0
\right\}
\\&=& \frac{4ac-b^2}{8a^{\frac{3}{2}}}
\left\{
\frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{a\left(ax^2+bx+c\right)}
+\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| }
-\ln{\sqrt{4ac-b^2}}
+C_0
\right\}
\\&=&\frac{4ac-b^2}{8a^{\frac{3}{2}}}\frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{\left(ax^2+bx+c\right)}
+\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| }
-\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{\sqrt{4ac-b^2}}
+\frac{4ac-b^2}{8a^{\frac{3}{2}}}C_0
\\&=&\frac{2ax+b}{4a}\sqrt{\left(ax^2+bx+c\right)}
+\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| }
-\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{\sqrt{4ac-b^2}}
+\frac{4ac-b^2}{8a^{\frac{3}{2}}}C_0
\\&=&\frac{2ax+b}{4a}\sqrt{\left(ax^2+bx+c\right)}
+\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| }
+C
\;\cdots\;C=-\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{\sqrt{4ac-b^2}}+\frac{4ac-b^2}{8a^{\frac{3}{2}}}C_0
\;(C:積分定数)
\end{eqnarray}
$$
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