[mathjax]
積分問題60番
計算
\(\displaystyle\int\displaystyle\frac{dx}{\sqrt{x^2+2x+2}}\)
\(\displaystyle\int\displaystyle\frac{dx}{\sqrt{(x+1)^2+1}}\)
ここで\(x+1=\tan \theta\) とおくと
\(\displaystyle\int\displaystyle\frac{d\theta}{\cos^2 \theta}\displaystyle\frac{1}{\sqrt{1+\tan^2 \theta}}\)
\(=\displaystyle\int\displaystyle\frac{d\theta}{\cos \theta}\)
\(=\displaystyle\frac{1}{2}\log\biggl|\displaystyle\frac{1+\sin \theta}{1-\sin \theta}\biggr|+C\) （この積分は有名な積分なので省略）
\(=\log\biggl|\displaystyle\frac{(1+\sin \theta)^2}{(1-\sin \theta)(1+\sin \theta)}\biggr|^{\frac{1}{2}}+C\)
\(=\log\biggl|\displaystyle\frac{1+\sin \theta}{\cos \theta}\biggr|+C\)
\(=\log|x+1+\sqrt{x^2+2x+2}|+C\) （変数を戻す）
答え
\(\log|x+1+\sqrt{x^2+2x+2}|+C\)
