【微分積分】4-4-2 部分分数分解の積分｜問題集

\(\displaystyle =\int\left(\frac{1}{x-2}-\frac{1}{x+5}\right)dx\)
\(\displaystyle =\log|x-2|-\log|x+5|+C\)
\(\displaystyle =\log\left|\frac{x-2}{x+5}\right|+C\)

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

\(\displaystyle =\int\left(1+\frac{3x+1}{x^2-3x+2}\right)dx\)
\(\displaystyle =\int\left(1+\frac{3x+1}{(x-2)(x-1)}\right)dx\)
\(\displaystyle =\int\left(1+\frac{7}{x-2}-\frac{4}{x-1}\right)dx\)
\(\displaystyle =x+7\log|x-2|-4\log|x-1|+C\)

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

\(\displaystyle I_n=\int\frac{1}{(x^2+16)^n}dx\)とおくと、
\(\displaystyle I_1=\int\frac{1}{x^2+16}dx\)
\(\displaystyle I_1=\frac{x}{x^2+16}+2\int\frac{x^2}{(x^2+16)^2}dx\)
\(\displaystyle I_1=\frac{x}{x^2+16}\)
\(\displaystyle \ \ \ +2\int\frac{1}{(x^2+16)^2}-32\int\frac{1}{(x^2+16)^2}dx\)
\(\displaystyle I_1=\frac{x}{x^2+16}+2I_1-32I_2\)
\(I_2\)について解く。
\(\displaystyle I_2=\frac{1}{32}\left\{\frac{x}{x^2+16}+I_1\right\}\)
\(\displaystyle =\frac{1}{128}\left\{\frac{4x}{x^2+16}+\tan^{-1}\frac{x}{4}\right\}+C\)
よって、
\(\displaystyle \int\frac{1}{(x^2+16)^2}dx=\frac{1}{128}\left\{\frac{4x}{x^2+16}+\tan^{-1}\frac{x}{4}\right\}+C\)

\(\displaystyle =\int\left(x^3+4x^2+12x+32+\frac{80x-128}{(x-2)^2}\right)dx\)
\(\displaystyle =\int\left(x^3+4x^2+12x+32+\frac{80}{x-2}+\frac{32}{(x-2)^2}\right)dx\)
\(\displaystyle =\frac{x^4}{4}+\frac{4x^3}{3}+6x^2+32x\)
\(\displaystyle \ \ \ +80\log|x-2|-\frac{32}{x-2}+C\)

\(t=x^3\)とおくと、\(\displaystyle dx=\frac{1}{3x^2}dt\)
\(\displaystyle =\frac{1}{3}\int\frac{t}{t^3-1}dt\)
\(\displaystyle =\frac{1}{9}\int\left(\frac{1}{t-1}+\frac{-t+1}{t^2+t+1}\right)dt\)
\(\displaystyle =\frac{1}{9}\log|t-1|-\frac{1}{18}\int\frac{2t+1}{t^2+t+1}dt\)
\(\displaystyle \ \ \ +\frac{1}{18}\int\frac{3}{(t+\frac{1}{2})^2+\frac{3}{4}}dt\)
\(\displaystyle =\frac{1}{9}\log|t-1|-\frac{1}{18}\log|t^2+t+1|\)
\(\displaystyle \ \ \ +2\sqrt{3}\tan^{-1}(\frac{2(t+\frac{1}{2})}{\sqrt{3}})+C\)
\(\displaystyle =\frac{1}{9}\log|x^3-1|-\frac{1}{18}\log|x^6+x^3+1|\)
\(\displaystyle \ \ \ +2\sqrt{3}\tan^{-1}(\frac{2x^3+1}{\sqrt{3}})+C\)

\(t=x^4\)とおくと、\(\displaystyle dx=\frac{1}{4x^3}dt\)
\(\displaystyle =\frac{1}{4}\int\frac{1}{t(t+1)}dt\)
\(\displaystyle =\frac{1}{4}\int\left(\frac{1}{t}-\frac{1}{t+1}\right)dt\)
\(\displaystyle =\frac{1}{4}(\log|t|-\log|t+1|)+C\)
\(\displaystyle =\frac{1}{4}(\log|x^4|-\log|x^4+1|)+C\)
\(\displaystyle =\frac{1}{4}\log\left|\frac{x^4}{x^4+1}\right|+C\)