【微分積分】4-4-1 有理関数の積分｜問題集

\(\displaystyle =\frac{1}{2}・\frac{1}{6}(2x+3)^6+C\)
\(\displaystyle =\frac{1}{12}(2x+3)^6+C\)

\(\displaystyle =\frac{1}{2}・\left(-\frac{1}{4}\right)(2x+3)^{-4}+C\)
\(\displaystyle =-\frac{1}{8(2x+3)^4}+C\)

\(\displaystyle =\int\frac{1}{(2x+2)^2+1}dx\)
\(\displaystyle =\frac{1}{2}\tan^{-1}(2x+2)+C\)

\(\displaystyle =\int\frac{1}{(\sqrt{2}x+\sqrt{2})^2+1}dx\)
\(\displaystyle =\frac{1}{\sqrt{2}}\tan^{-1}\sqrt{2}(x+1)+C\)

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

\(\displaystyle =-\frac{1}{3}・\left(-\frac{1}{2}\right)(2-3x)^{-2}+C\)
\(\displaystyle =\frac{1}{6(2-3x)^2}+C\)

\(\displaystyle =\int\frac{1}{(x-1)^2+1}dx\)
\(\displaystyle =\tan^{-1}(x-1)+C\)

\(t=x^3\)とおくと、\(\displaystyle dx=\frac{1}{3x^2}dt\)
\(\displaystyle =\frac{1}{3}\int\frac{1}{t^2+1}dt\)
\(\displaystyle =\frac{1}{3}\tan^{-1}t+C\)
\(\displaystyle =\frac{1}{3}\tan^{-1}x^3+C\)

\(t=x^4\)とおくと、\(\displaystyle dx=\frac{1}{4x^3}dt\)
\(\displaystyle =\frac{1}{4}\int\frac{1}{t^2+1}dt\)
\(\displaystyle =\frac{1}{4}\tan^{-1}t+C\)
\(\displaystyle =\frac{1}{4}\tan^{-1}x^4+C\)