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積分問題6番
思考
分子分母とも\( (x-1) \) で割れるので割る。
計算
\(\displaystyle\int\displaystyle\frac{x^5-x^4+x^3-2x^2+x}{x^6-1}dx \)
\(=\displaystyle\int\displaystyle\frac{x^4+x^2-x}{(x+1)(x^4+x^2+1)}dx \) \(\cdots\) 分子分母を\((x-1)\)で割る
\(=\displaystyle\int\biggl(\displaystyle\frac{1}{x+1}-\displaystyle\frac{1}{x^4+x^2+1}\biggr)dx\) \(\cdots\) 部分分数分解
\(=\log|x+1| -\)\(\displaystyle\int\displaystyle\frac{dx}{x^4+x^2+1}\)
第2項
\( x^4+x^2+1=(x^2+x+1)(x^2-x+1)\) なので部分分数分解すると
\(\displaystyle\int\displaystyle\frac{dx}{x^4+x^2+1}\)
\(=\displaystyle\int\displaystyle\frac{dx}{(x^2+x+1)(x^2-x+1)}\)
\(= -\displaystyle\frac{1}{2} \displaystyle\int\displaystyle\frac{x-1}{x^2-x+1}dx+\displaystyle\frac{1}{2} \displaystyle\int\displaystyle\frac{x+1}{x^2+x+1}dx\)
\(=-\displaystyle\frac{1}{4} \displaystyle\int\displaystyle\frac{(2x-1)-1}{x^2-x+1}dx+\displaystyle\frac{1}{4} \displaystyle\int\displaystyle\frac{(2x+1)+1}{x^2+x+1}dx\)
\(=-\displaystyle\frac{1}{4} \displaystyle\int\displaystyle\frac{2x-1}{x^2-x+1}dx+\displaystyle\frac{1}{4} \displaystyle\int\displaystyle\frac{dx}{x^2-x+1}\)
\(+\displaystyle\frac{1}{4} \displaystyle\int\displaystyle\frac{2x+1}{x^2+x+1}dx+\displaystyle\frac{1}{4}\displaystyle\int\displaystyle\frac{dx}{x^2+x+1}\)
\(=\displaystyle\frac{1}{4}\log\biggl(\displaystyle\frac{x^2+x+1}{x^2-x+1}\biggr)+\)\(\displaystyle\frac{1}{4}\displaystyle\int\displaystyle\frac{dx}{x^2-x+1}\)\(+\displaystyle\frac{1}{4} \displaystyle\int\displaystyle\frac{dx}{x^2+x+1}\)
後ろの二項を考える。
\(\displaystyle\int\displaystyle\frac{dx}{(x-a)^2+b^2}=\displaystyle\frac{1}{b}\tan^{-1} \biggl(\displaystyle\frac{x-a}{b}\biggr)+C \)
という式を使う。※\(x-a=b\tan \theta\) と置き換えればわかります。
詳しくは下記の記事を参照ください。
第二項
\(\displaystyle\frac{1}{4}\displaystyle\int\displaystyle\frac{dx}{x^2-x+1}\)
\(= \displaystyle\frac{1}{4}\displaystyle\int\displaystyle\frac{dx}{\biggl(x-\displaystyle\frac{1}{2}\biggr)^2+\biggl(\displaystyle\frac{\sqrt3}{2}\biggr)^2}\)
\(=\displaystyle\frac{1}{2\sqrt3}\tan^{-1} \biggl(\displaystyle\frac{2x-1}{\sqrt3}\biggr)+C\)
第三項
\(\displaystyle\frac{1}{4}\displaystyle\int\displaystyle\frac{dx}{x^2+x+1}\)
\(= \displaystyle\frac{1}{4}\displaystyle\int\displaystyle\frac{dx}{\biggl(x+\displaystyle\frac{1}{2}\biggr)^2+\biggl(\displaystyle\frac{\sqrt3}{2}\biggr)^2}\)
\(=\displaystyle\frac{1}{2\sqrt3}\tan^{-1} \biggl(\displaystyle\frac{2x+1}{\sqrt3}\biggr)+C\)
第二項の結果
\(\displaystyle\int\displaystyle\frac{dx}{x^4+x^2+1}\)
\(=\displaystyle\frac{1}{4}\log\biggl(\displaystyle\frac{x^2+x+1}{x^2-x+1}\biggr)+\displaystyle\frac{1}{2\sqrt3}\tan^{-1} \biggl(\displaystyle\frac{2x-1}{\sqrt3}\biggr)+\displaystyle\frac{1}{2\sqrt3}\tan^{-1} \biggl(\displaystyle\frac{2x+1}{\sqrt3}\biggr)+C\)
答え
よって、問題の積分の答えは以下のようになる。
\(\log|x+1| -\displaystyle\frac{1}{4}\log\biggl(\displaystyle\frac{x^2+x+1}{x^2-x+1}\biggr)-\displaystyle\frac{1}{2\sqrt3}\tan^{-1} \biggl(\displaystyle\frac{2x-1}{\sqrt3}\biggr)-\displaystyle\frac{1}{2\sqrt3}\tan^{-1} \biggl(\displaystyle\frac{2x+1}{\sqrt3}\biggr)+C\)
