【微分積分】4-4-2 部分分数分解の積分｜要点まとめ

\(\displaystyle =x+\frac{x}{x^3-1}\)
\(\displaystyle =x+\frac{x}{(x-1)(x^2+x+1)}\)
ここで、
\(\displaystyle \frac{x}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}\)
とおくと、
\(\displaystyle A=(x-1)\frac{x}{x^3-1}\Big|_{x=1}=\frac{1}{3}\)
\(\displaystyle Bx+C=(x^2+x+1)\frac{x}{x^3-1}\Big|_{x=\frac{-1\pm\sqrt{3}i}{2}}\)
\(\displaystyle B・\frac{-1\pm\sqrt{3}i}{2}+C=\frac{6-2\sqrt{3}i}{12}\)
\(\displaystyle A=\frac{1}{3},B=-\frac{1}{3},C=\frac{1}{3}\)
よって、
\(\displaystyle \frac{x^4}{x^3-1}=x+\frac{1}{3(x-1)}-\frac{x-1}{3(x^2+x+1)}\)

\(\displaystyle \frac{3x^2-1}{x^3(x^2+1)^2}\)
\(\displaystyle \ \ \ =\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{Dx+E}{x^2+1}+\frac{Fx+G}{(x^2+1)^2}\)
とおくと、
\(\displaystyle C=x^3\frac{3x^2-1}{x^3(x^2+1)^2}\Big|_{x=0}=-1\)
\(\displaystyle B=\frac{d}{dx}x^3\frac{3x^2-1}{x^3(x^2+1)^2}\Big|_{x=0}=0\)
\(\displaystyle A=\frac{1}{2!}\frac{d^2}{dx^2}x^3\frac{3x^2-1}{x^3(x^2+1)^2}\Big|_{x=0}=5\)
\(\displaystyle Fx+G=(x^2+1)^2\frac{3x^2-1}{x^3(x^2+1)^2}\Big|_{x=i}\)
\(F=-4,G=0\)
\(\displaystyle Dx+E=\frac{d}{dx}(x^2+1)^2\frac{3x^2-1}{x^3(x^2+1)^2}\Big|_{x=i}\)
\(D=-5,E=0\)
よって、
\(\displaystyle \frac{3x^2-1}{x^3(x^2+1)^2}\)
\(\displaystyle \ \ \ =\frac{5}{x}-\frac{1}{x^3}-\frac{5x}{x^2+1}-\frac{4x}{(x^2+1)^2}\)

\(\displaystyle =\int\left(\frac{1}{x}-\frac{1}{x+1}\right)dx\)
\(\displaystyle =\log|x|-\log|x+1|+C\)
\(\displaystyle =\log\left|\frac{x}{x+1}\right|+C\)

\(\displaystyle =\int\left(-\frac{1}{2x}+\frac{1}{x+1}+\frac{1}{2(x+2)}\right)dx\)
\(\displaystyle =-\frac{1}{2}\log|x|+\log|x+1|+\frac{1}{2}\log|x+2|+C\)
\(\displaystyle =\frac{1}{2}\log\frac{(x+1)^2|x+2|}{|x|}+C\)

\(\displaystyle =\int\left(-\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x+1}\right)dx\)
\(\displaystyle =-\log|x|-\frac{1}{x}+\log|x+1|+C\)
\(\displaystyle =\log\left|\frac{x+1}{x}\right|-\frac{1}{x}+C\)

\(\displaystyle =\int\left(x+1-\frac{3}{x^3-1}\right)dx\)
\(\displaystyle =\int\left(x+1-\frac{3}{(x-1)(x^2+x+1)}\right)dx\)
\(\displaystyle =\int\left(x+1-\frac{1}{x-1}+\frac{x+2}{x^2+x+1}\right)dx\)
\(\displaystyle =\frac{x^2}{2}+x-\log|x-1|+\int\frac{x+2}{x^2+x+1}dx\)
\(\displaystyle =\frac{x^2}{2}+x-\log|x-1|+\frac{1}{2}\int\frac{2x}{x^2+x+1}dx\)
\(\displaystyle \ \ \ +\frac{3}{2}\int\frac{1}{x^2+x+1}dx\)
\(\displaystyle =\frac{x^2}{2}+x-\log|x-1|\)
\(\displaystyle \ \ \ +\frac{1}{2}\int\frac{(x^2+x+1)'}{x^2+x+1}dx\)
\(\displaystyle \ \ \ +\frac{3}{2}\int\frac{1}{(x+\frac{1}{2})^2+\frac{3}{4}}dx\)
\(\displaystyle =\frac{x^2}{2}+x-\log|x-1|+\frac{1}{2}\log|x^2+x+1|\)
\(\displaystyle \ \ \ +\frac{3}{2}・\frac{2}{\sqrt{3}}\tan^{-1}\frac{2}{\sqrt{3}}\left(x+\frac{1}{2}\right)+C\)
\(\displaystyle =\frac{x^2}{2}+x+\frac{1}{2}\log\frac{x^2+x+1}{(x-1)^2}\)
\(\displaystyle \ \ \ +\sqrt{3}\tan^{-1}\frac{2x+1}{\sqrt{3}}+C\)

\(\displaystyle =\int\left(x^3-1+\frac{2}{x^3+1}\right)dx\)
\(\displaystyle =\int\left(x^3-1+\frac{2}{(x+1)(x^2-x+1)}\right)dx\)
\(\displaystyle =\int\left(x^3-1+\frac{2}{3}\frac{1}{x+1}+\frac{2}{3}\frac{-x+2}{x^2-x+1}\right)dx\)
\(\displaystyle =\frac{x^4}{4}-x+\frac{2}{3}\log|x+1|\)
\(\displaystyle \ \ \ +\frac{2}{3}\int\frac{-x+2}{x^2-x+1}dx\)
\(\displaystyle =\frac{x^4}{4}-x+\frac{2}{3}\log|x+1|\)
\(\displaystyle \ \ \ -\frac{1}{3}\int\frac{2x-1}{x^2-x+1}dx\)
\(\displaystyle \ \ \ +\int\frac{1}{x^2-x+1}dx\)
\(\displaystyle =\frac{x^4}{4}-x+\frac{2}{3}\log|x+1|\)
\(\displaystyle \ \ \ -\frac{1}{3}\int\frac{(x^2-x+1)'}{x^2-x+1}dx\)
\(\displaystyle \ \ \ +\int\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}dx\)
\(\displaystyle =\frac{x^4}{4}-x+\frac{2}{3}\log|x+1|-\frac{1}{2}\log|x^2-x+1|\)
\(\displaystyle \ \ \ +\frac{2}{\sqrt{3}}\tan^{-1}\frac{2}{\sqrt{3}}\left(x-\frac{1}{2}\right)+C\)
\(\displaystyle =\frac{x^4}{4}-x+\frac{1}{3}\log\frac{(x+1)^2}{x^2-x+1}\)
\(\displaystyle \ \ \ +\frac{2}{\sqrt{3}}\tan^{-1}\frac{2x-1}{\sqrt{3}}+C\)