integral of (x^2)/((x^2+4x+5)^2)

解

\int \frac{x ^{2}}{( x ^{2} + 4 x + 5 ) ^{2}} d x

- x sin (2 arctan (x + 2)) + \frac{5}{2} arctan (x + 2) - \frac{5}{4} sin (2 arctan (x + 2)) + C

解答ステップ

\int \frac{x ^{2}}{( x ^{2} + 4 x + 5 ) ^{2}} d x

以下の部分分数を得る:

\frac{x ^{2}}{( x ^{2} + 4 x + 5 ) ^{2}} : \frac{1}{x ^{2} + 4 x + 5} + \frac{- 4 x - 5}{( x ^{2} + 4 x + 5 ) ^{2}}

= \int \frac{1}{x ^{2} + 4 x + 5} + \frac{- 4 x - 5}{( x ^{2} + 4 x + 5 ) ^{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{1}{x ^{2} + 4 x + 5} d x + \int \frac{- 4 x - 5}{( x ^{2} + 4 x + 5 ) ^{2}} d x

\int \frac{1}{x ^{2} + 4 x + 5} d x = arctan (x + 2)

\int \frac{- 4 x - 5}{( x ^{2} + 4 x + 5 ) ^{2}} d x = - x sin (2 arctan (x + 2)) + \frac{3}{2} arctan (x + 2) - \frac{5}{4} sin (2 arctan (x + 2))

= arctan (x + 2) - x sin (2 arctan (x + 2)) + \frac{3}{2} arctan (x + 2) - \frac{5}{4} sin (2 arctan (x + 2))

簡素化

arctan (x + 2) - x sin (2 arctan (x + 2)) + \frac{3}{2} arctan (x + 2) - \frac{5}{4} sin (2 arctan (x + 2)) : - x sin (2 arctan (x + 2)) + \frac{5}{2} arctan (x + 2) - \frac{5}{4} sin (2 arctan (x + 2))

= - x sin (2 arctan (x + 2)) + \frac{5}{2} arctan (x + 2) - \frac{5}{4} sin (2 arctan (x + 2))

定数を解答に追加する

= - x sin (2 arctan (x + 2)) + \frac{5}{2} arctan (x + 2) - \frac{5}{4} sin (2 arctan (x + 2)) + C