integral of ((8x^2+9x+8))/((x^2+1)^2)

解

\int \frac{( 8 x ^{2} + 9 x + 8 )}{( x ^{2} + 1 ) ^{2}} d x

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

解答ステップ

\int \frac{8 x ^{2} + 9 x + 8}{( x ^{2} + 1 ) ^{2}} d x

以下の部分分数を得る:

\frac{8 x ^{2} + 9 x + 8}{( x ^{2} + 1 ) ^{2}} : \frac{8}{x ^{2} + 1} + \frac{9 x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{8}{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{8}{x ^{2} + 1} d x + \int \frac{9 x}{( x ^{2} + 1 ) ^{2}} d x

\int \frac{8}{x ^{2} + 1} d x = 8 arctan (x)

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

= 8 arctan (x) - \frac{9}{2 ( x ^{2} + 1 )}

定数を解答に追加する

= 8 arctan (x) - \frac{9}{2 ( x ^{2} + 1 )} + C