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
$$\begin{eqnarray}
\left.\int \frac{1}{x^2+px-1}\mathrm{d}x\right|_{p=0}
&=&\left.\frac{1}{\sqrt{p^2+4}}\log{\frac{
\left|x+\frac{p}{2}-\sqrt{\frac{p^2}{4}+1}\right| }{
\left|x+\frac{p}{2}+\sqrt{\frac{p^2}{4}+1}\right| }}\right|_{p=0}
\\&=&\frac{1}{\sqrt{0^2+4}}\log{\frac{
\left|x+\frac{0}{2}-\sqrt{\frac{0^2}{4}+1}\right| }{
\left|x+\frac{0}{2}+\sqrt{\frac{0^2}{4}+1}\right| }}
\\&=&\frac{1}{2}\log{\frac{
\left|x-1\right| }{
\left|x+1\right| }} + C\;\;(C:積分定数)
\end{eqnarray}$$
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