[mathjax]
積分問題25番
思考
分母を因数分解して部分分数分解します。
言うのは簡単ですが、計算は大変です。
計算
※\(x^6+1=(x^2+1)(x^4-x^2+1)\)
\(\displaystyle\int\displaystyle\frac{dx}{(x^2+1)(x^4-x^2+1)}\)
\(=\displaystyle\frac{1}{3}\displaystyle\int\displaystyle\frac{dx}{x^2+1}\)\(-\displaystyle\frac{1}{3}\displaystyle\int\displaystyle\frac{x^2-2}{x^4-x^2+1}dx\)
第1項
\(=\displaystyle\frac{1}{3}\displaystyle\int\displaystyle\frac{dx}{x^2+1}=\displaystyle\frac{1}{3}\tan^{-1} x+C\)
第2項
\(\displaystyle\int\displaystyle\frac{x^2-2}{x^4-x^2+1}dx\)
\(=\displaystyle\frac{\sqrt3}{6}\displaystyle\int\displaystyle\frac{3x-2\sqrt3}{x^2-\sqrt3 x+1}dx-\displaystyle\frac{\sqrt3}{6}\displaystyle\int\displaystyle\frac{3x+2\sqrt3}{x^2+\sqrt3 x+1}dx\)
\(=\displaystyle\frac{\sqrt3}{4}\displaystyle\int\displaystyle\frac{2x-\sqrt3}{x^2-\sqrt3 x+1}dx-\displaystyle\frac{1}{4}\displaystyle\int\displaystyle\frac{dx}{\biggl(x-\displaystyle\frac{\sqrt3}{2}\biggr)^2+\biggl(\displaystyle\frac{1}{2}\biggr)^2}\)
\(-\displaystyle\frac{\sqrt3}{4}\displaystyle\int\displaystyle\frac{2x+\sqrt3}{x^2+\sqrt3 x+1}dx-\displaystyle\frac{1}{4}\displaystyle\int\displaystyle\frac{dx}{\biggl(x+\displaystyle\frac{\sqrt3}{2}\biggr)^2+\biggl(\displaystyle\frac{1}{2}\biggr)^2}\)
\(=\displaystyle\frac{\sqrt3}{4}\log\biggl(\displaystyle\frac{x^2-\sqrt3x+1}{x^2+\sqrt3 x+1}\biggr)-\displaystyle\frac{1}{2}\tan^{-1}(2x-\sqrt3)-\displaystyle\frac{1}{2}\tan^{-1}(2x+\sqrt3)+C\)
答え
まとめると答えになる。
\(\displaystyle\frac{1}{3}\tan^{-1} x-\displaystyle\frac{\sqrt3}{12}\log\biggl(\displaystyle\frac{x^2-\sqrt3x+1}{x^2+\sqrt3 x+1}\biggr)\)
\(+\displaystyle\frac{1}{6}\tan^{-1}(2x-\sqrt3)+\displaystyle\frac{1}{6}\tan^{-1}(2x+\sqrt3)+C\)
