integral of (2x^2+9x+2)/((x^2+1)^2)

解

\int \frac{2 x ^{2} + 9 x + 2}{( x ^{2} + 1 ) ^{2}} d x

2 arctan (x) - \frac{9}{2 ( x ^{2} + 1 )} + C

解答ステップ

\int \frac{2 x ^{2} + 9 x + 2}{( x ^{2} + 1 ) ^{2}} d x

以下の部分分数を得る:

\frac{2 x ^{2} + 9 x + 2}{( x ^{2} + 1 ) ^{2}} : \frac{2}{x ^{2} + 1} + \frac{9 x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{2}{x ^{2} + 1} + \frac{9 x}{( 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 \frac{2}{x ^{2} + 1} d x + \int \frac{9 x}{( x ^{2} + 1 ) ^{2}} d x

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

\int \frac{9 x}{( x ^{2} + 1 ) ^{2}} d x = - \frac{9}{2 ( x ^{2} + 1 )}

= 2 arctan (x) - \frac{9}{2 ( x ^{2} + 1 )}

定数を解答に追加する

= 2 arctan (x) - \frac{9}{2 ( x ^{2} + 1 )} + C