1/(x^2+px-1)の不定積分

1/(x^2+px-1)の不定積分

$$\begin{array}{r}\int \frac{1}{x^2+px-1}\mathrm{d}x\dots p\mathrm{は}\mathrm{実}\mathrm{数}\end{array}$$

---

1/(x^2+px-1)の部分分数分解

$$\begin{array}{rcl}\frac{1}{x^2+px-1}& =& \frac{1}{x^2+2\frac{p}{2}x+{\left(\frac{p}{2}\right)}^2-{\left(\frac{p}{2}\right)}^2-1}\\ & =& \frac{1}{{(x+\frac{p}{2})}^2-(\frac{p^2}{4}+1)}\\ & =& \frac{1}{{(x+\frac{p}{2})}^2-{\left(\sqrt{\frac{p^2}{4}+1}\right)}^2}\dots \frac{p^2}{4}+1>0\mathrm{な}\mathrm{の}\mathrm{で}\mathrm{問}\mathrm{題}\mathrm{な}\mathrm{く}\mathrm{根}\mathrm{号}\mathrm{を}\mathrm{つ}\mathrm{け}\mathrm{ら}\mathrm{れ}\mathrm{る}\\ & =& \frac{1}{(x+\frac{p}{2}+\sqrt{\frac{p^2}{4}+1})(x+\frac{p}{2}-\sqrt{\frac{p^2}{4}+1})}\\ & =& \frac{1}{{\lambda }_1{\lambda }_2}\dots {\lambda }_{1,2}=x+\frac{p}{2}\pm \sqrt{\frac{p^2}{4}+1}\\ & =& \frac{C_1}{{\lambda }_1}+\frac{C_2}{{\lambda }_2}=\frac{C_1{\lambda }_2+C_2{\lambda }_1}{{\lambda }_1{\lambda }_2}\dots \mathrm{部}\mathrm{分}\mathrm{分}\mathrm{数}\mathrm{分}\mathrm{解}\\ & =& \frac{\frac{-1}{\sqrt{p^2+4}}}{{\lambda }_1}+\frac{\frac{1}{\sqrt{p^2+4}}}{{\lambda }_2}\\ & & \dots C_1{\lambda }_2+C_2{\lambda }_1=1\\ & & \dots C_{1}x+C_1(\frac{p}{2}-\sqrt{\frac{p^2}{4}+1})+C_{2}x+C_2(\frac{p}{2}+\sqrt{\frac{p^2}{4}+1})=1\\ & & \dots \{\begin{array}{}(C_1+C_2)x=0\\ C_1(\frac{p}{2}-\sqrt{\frac{p^2}{4}+1})+C_2(\frac{p}{2}+\sqrt{\frac{p^2}{4}+1})=1\end{array}\dots \mathrm{係}\mathrm{数}\mathrm{比}\mathrm{較}\\ & & \dots C_1=-C_2\dots \mathrm{一}\mathrm{つ}\mathrm{目}\mathrm{の}\mathrm{式}\mathrm{よ}\mathrm{り}\\ & & \dots C_1(\frac{p}{2}-\sqrt{\frac{p^2}{4}+1})-C_1(\frac{p}{2}+\sqrt{\frac{p^2}{4}+1})=1\\ & & \dots C_1\{(\frac{p}{2}-\sqrt{\frac{p^2}{4}+1})-(\frac{p}{2}+\sqrt{\frac{p^2}{4}+1})\}=1\\ & & \dots -2C_1\sqrt{\frac{p^2}{4}+1}=1\\ & & \dots -C_1\sqrt{p^2+4}=1\\ & & \dots C_1=\frac{-1}{\sqrt{p^2+4}}\\ & & \dots C_2=-C_1=\frac{1}{\sqrt{p^2+4}}\\ & =& \frac{1}{\sqrt{p^2+4}}(\frac{-1}{{\lambda }_1}+\frac{1}{{\lambda }_2})\\ & =& \frac{1}{\sqrt{p^2+4}}(\frac{-1}{x+\frac{p}{2}+\sqrt{\frac{p^2}{4}+1}}+\frac{1}{x+\frac{p}{2}-\sqrt{\frac{p^2}{4}+1}})\end{array}$$

---

1/(x^2+px-1)の不定積分 計算

$$\begin{array}{rcl}\int \frac{1}{x^2+px-1}\mathrm{d}x& =& \int \frac{1}{\sqrt{p^2+4}}(\frac{-1}{x+\frac{p}{2}+\sqrt{\frac{p^2}{4}+1}}+\frac{1}{x+\frac{p}{2}-\sqrt{\frac{p^2}{4}+1}})\mathrm{d}x\\ & =& \frac{1}{\sqrt{p^2+4}}(\int \frac{-1}{x+\frac{p}{2}+\sqrt{\frac{p^2}{4}+1}}\mathrm{d}x+\int \frac{1}{x+\frac{p}{2}-\sqrt{\frac{p^2}{4}+1}}\mathrm{d}x)\\ & =& \frac{1}{\sqrt{p^2+4}}(\int \frac{-1}{u}\mathrm{d}u+\int \frac{1}{v}\mathrm{d}v)\\ & & \dots u=x+\frac{p}{2}+\sqrt{\frac{p^2}{4}+1},\frac{\mathrm{d}u}{\mathrm{d}x}=1,\mathrm{d}x=\mathrm{d}u\\ & & \dots v=x+\frac{p}{2}-\sqrt{\frac{p^2}{4}+1},\frac{\mathrm{d}v}{\mathrm{d}x}=1,\mathrm{d}x=\mathrm{d}v\\ & =& \frac{1}{\sqrt{p^2+4}}(-\log \left|u\right|+\log \left|v\right|)\\ & & \dots \int \frac{1}{x}\mathrm{d}x=\log \left|x\right|+C(C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\\ & =& \frac{1}{\sqrt{p^2+4}}(-\log |x+\frac{p}{2}+\sqrt{\frac{p^2}{4}+1}|+\log |x+\frac{p}{2}-\sqrt{\frac{p^2}{4}+1}|)\\ & =& \frac{1}{\sqrt{p^2+4}}\log \frac{|x+\frac{p}{2}-\sqrt{\frac{p^2}{4}+1}|}{|x+\frac{p}{2}+\sqrt{\frac{p^2}{4}+1}|}+C(C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\\ & & \dots \log A-\log B=\log \frac{A}{B}\end{array}$$

---

$p=0$の場合

$$\begin{array}{rcl}{\int \frac{1}{x^2+px-1}\mathrm{d}x|}_{p=0}& =& {\frac{1}{\sqrt{p^2+4}}\log \frac{|x+\frac{p}{2}-\sqrt{\frac{p^2}{4}+1}|}{|x+\frac{p}{2}+\sqrt{\frac{p^2}{4}+1}|}|}_{p=0}\\ & =& \frac{1}{\sqrt{0^2+4}}\log \frac{|x+\frac{0}{2}-\sqrt{\frac{0^2}{4}+1}|}{|x+\frac{0}{2}+\sqrt{\frac{0^2}{4}+1}|}\\ & =& \frac{1}{2}\log \frac{|x-1|}{|x+1|}+C(C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\end{array}$$