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

解

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

x - \frac{3}{2} arctan (x) + \frac{x}{2 ( 1 + x ^{2} )} + C

解答ステップ

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

以下の部分分数を得る:

\frac{x ^{4}}{( x ^{2} + 1 ) ^{2}} : 1 - \frac{2}{x ^{2} + 1} + \frac{1}{( x ^{2} + 1 ) ^{2}}

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

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

\int 1 d x = x

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

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

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

簡素化

x - 2 arctan (x) + \frac{1}{2} (arctan (x) + \frac{x}{1 + x ^{2}}) : x - \frac{3}{2} arctan (x) + \frac{x}{2 ( 1 + x ^{2} )}

= x - \frac{3}{2} arctan (x) + \frac{x}{2 ( 1 + x ^{2} )}

定数を解答に追加する

= x - \frac{3}{2} arctan (x) + \frac{x}{2 ( 1 + x ^{2} )} + C