[mathjax]
積分問題9番
思考
根号が出てはきますが、分母を因数分解して部分分数分解して計算していきます。
計算
\(\displaystyle\int\displaystyle\frac{dx}{x^4+1}\)
\(=\displaystyle\int\displaystyle\frac{dx}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}+1)}\) （因数分解）
\(=-\displaystyle\frac{1}{4}\displaystyle\int\displaystyle\frac{\sqrt{2}x-2}{x^2-\sqrt{2}x+1}dx+\displaystyle\frac{1}{4}\displaystyle\int\displaystyle\frac{\sqrt{2}x+2}{x^2+\sqrt{2}x+1}dx\) （部分分数分解）
\(=-\displaystyle\frac{\sqrt2}{8}\displaystyle\int\displaystyle\frac{(2x-\sqrt2)-\sqrt2}{x^2-\sqrt{2}x+1}dx+\displaystyle\frac{\sqrt2}{8}\displaystyle\int\displaystyle\frac{(2x+\sqrt2)+\sqrt2}{x^2+\sqrt{2}x+1}dx\)
\(=-\displaystyle\frac{\sqrt2}{8}\displaystyle\int\displaystyle\frac{2x-\sqrt2}{x^2-\sqrt{2}x+1}dx\)\(+\displaystyle\frac{1}{4}\displaystyle\int\displaystyle\frac{dx}{\biggl(x-\displaystyle\frac{\sqrt2}{2}\biggr)^2+\biggl(\displaystyle\frac{\sqrt2}{2}\biggr)^2}\)
\(+\displaystyle\frac{\sqrt2}{8}\displaystyle\int\displaystyle\frac{2x+\sqrt2}{x^2+\sqrt{2}x+1}dx+\) \(\displaystyle\frac{1}{4}\displaystyle\int\displaystyle\frac{dx}{\biggl(x+\displaystyle\frac{\sqrt2}{2}\biggr)^2+\biggl(\displaystyle\frac{\sqrt2}{2}\biggr)^2}\)
\(=\displaystyle\frac{\sqrt2}{8}\log\biggl(\displaystyle\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\biggr)+\displaystyle\frac{\sqrt2}{4}\tan^{-1}(\sqrt{2}x-1)+\displaystyle\frac{\sqrt2}{4}\tan^{-1}(\sqrt{2}x+1)+C\)
※最後の変形で \(\tan^{-1}x\) が登場する積分に関しては以下の事実を使用しました。
答え
\(=\displaystyle\frac{\sqrt2}{8}\log\biggl(\displaystyle\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\biggr)+\displaystyle\frac{\sqrt2}{4}\tan^{-1}(\sqrt{2}x-1)+\displaystyle\frac{\sqrt2}{4}\tan^{-1}(\sqrt{2}x+1)+C\)
