integral of (3x^2-8x+5)/((x+1)^4)

解

\int \frac{3 x ^{2} - 8 x + 5}{( x + 1 ) ^{4}} d x

- \frac{x ^{2}}{( x + 1 ) ^{3}} + \frac{- 2 x - 1}{( x + 1 ) ^{2}} + \frac{4}{( x + 1 ) ^{2}} - \frac{13}{3 ( x + 1 ) ^{3}} + C

解答ステップ

\int \frac{3 x ^{2} - 8 x + 5}{( x + 1 ) ^{4}} d x

拡張

\frac{3 x ^{2} - 8 x + 5}{( x + 1 ) ^{4}} : \frac{3 x ^{2}}{( x + 1 ) ^{4}} - \frac{8 x}{( x + 1 ) ^{4}} + \frac{5}{( x + 1 ) ^{4}}

= \int \frac{3 x ^{2}}{( x + 1 ) ^{4}} - \frac{8 x}{( x + 1 ) ^{4}} + \frac{5}{( x + 1 ) ^{4}} 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{3 x ^{2}}{( x + 1 ) ^{4}} d x - \int \frac{8 x}{( x + 1 ) ^{4}} d x + \int \frac{5}{( x + 1 ) ^{4}} d x

\int \frac{3 x ^{2}}{( x + 1 ) ^{4}} d x = - \frac{x ^{2}}{( x + 1 ) ^{3}} + \frac{- 2 x - 1}{( x + 1 ) ^{2}}

\int \frac{8 x}{( x + 1 ) ^{4}} d x = 8 (- \frac{1}{2 ( x + 1 ) ^{2}} + \frac{1}{3 ( x + 1 ) ^{3}})

\int \frac{5}{( x + 1 ) ^{4}} d x = - \frac{5}{3 ( x + 1 ) ^{3}}

= - \frac{x ^{2}}{( x + 1 ) ^{3}} + \frac{- 2 x - 1}{( x + 1 ) ^{2}} - 8 (- \frac{1}{2 ( x + 1 ) ^{2}} + \frac{1}{3 ( x + 1 ) ^{3}}) - \frac{5}{3 ( x + 1 ) ^{3}}

簡素化

- \frac{x ^{2}}{( x + 1 ) ^{3}} + \frac{- 2 x - 1}{( x + 1 ) ^{2}} - 8 (- \frac{1}{2 ( x + 1 ) ^{2}} + \frac{1}{3 ( x + 1 ) ^{3}}) - \frac{5}{3 ( x + 1 ) ^{3}} : - \frac{x ^{2}}{( x + 1 ) ^{3}} + \frac{- 2 x - 1}{( x + 1 ) ^{2}} + \frac{4}{( x + 1 ) ^{2}} - \frac{13}{3 ( x + 1 ) ^{3}}

= - \frac{x ^{2}}{( x + 1 ) ^{3}} + \frac{- 2 x - 1}{( x + 1 ) ^{2}} + \frac{4}{( x + 1 ) ^{2}} - \frac{13}{3 ( x + 1 ) ^{3}}

定数を解答に追加する

= - \frac{x ^{2}}{( x + 1 ) ^{3}} + \frac{- 2 x - 1}{( x + 1 ) ^{2}} + \frac{4}{( x + 1 ) ^{2}} - \frac{13}{3 ( x + 1 ) ^{3}} + C