integral of (x^2+2x+1)/(x^4+2x^2+1)

解

\int \frac{x ^{2} + 2 x + 1}{x ^{4} + 2 x ^{2} + 1} d x

arctan (x) - \frac{1}{x ^{2} + 1} + C

解答ステップ

\int \frac{x ^{2} + 2 x + 1}{x ^{4} + 2 x ^{2} + 1} d x

拡張

\frac{x ^{2} + 2 x + 1}{x ^{4} + 2 x ^{2} + 1} : \frac{x ^{2}}{x ^{4} + 2 x ^{2} + 1} + \frac{2 x}{x ^{4} + 2 x ^{2} + 1} + \frac{1}{x ^{4} + 2 x ^{2} + 1}

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x ^{2}}{x ^{4} + 2 x ^{2} + 1} d x + \int \frac{2 x}{x ^{4} + 2 x ^{2} + 1} d x + \int \frac{1}{x ^{4} + 2 x ^{2} + 1} d x

\int \frac{x ^{2}}{x ^{4} + 2 x ^{2} + 1} d x = \frac{1}{2} arctan (x) - \frac{x}{2 ( 1 + x ^{2} )}

\int \frac{2 x}{x ^{4} + 2 x ^{2} + 1} d x = - \frac{1}{x ^{2} + 1}

\int \frac{1}{x ^{4} + 2 x ^{2} + 1} d x = \frac{1}{2} (arctan (x) + \frac{x}{1 + x ^{2}})

= \frac{1}{2} arctan (x) - \frac{x}{2 ( 1 + x ^{2} )} - \frac{1}{x ^{2} + 1} + \frac{1}{2} (arctan (x) + \frac{x}{1 + x ^{2}})

簡素化

\frac{1}{2} arctan (x) - \frac{x}{2 ( 1 + x ^{2} )} - \frac{1}{x ^{2} + 1} + \frac{1}{2} (arctan (x) + \frac{x}{1 + x ^{2}}) : arctan (x) - \frac{1}{x ^{2} + 1}

= arctan (x) - \frac{1}{x ^{2} + 1}

定数を解答に追加する

= arctan (x) - \frac{1}{x ^{2} + 1} + C