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

解答

\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

求解步骤

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

乘开

\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

使用积分加法定则:

\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}}

化简

- \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}}

解答补常数

= - \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

(\frac{1}{1 - x})^{^{'}}

l a pl a ce t r an s f or m e^{2 t} sin (j^{2} t)

\frac{\partial}{\partial x} ((9 + 20 x) y^{t a n (x)})

t an g e n t f (x) = - x^{2}, (- 3, - 9)

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