integral of (x-2)/((x^2+2x+2)^2)

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

簡素化

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

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

定数を解答に追加する

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