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受欢迎的微积分 >

integral of t/((t+1)^4)

解答

\int \frac{t}{( t + 1 ) ^{4}} d t

解答

- \frac{1}{2 ( t + 1 ) ^{2}} + \frac{1}{3 ( t + 1 ) ^{3}} + C

求解步骤

\int \frac{t}{( t + 1 ) ^{4}} d t

使用换元积分法

= \int \frac{u - 1}{u ^{4}} d u

乘开

\frac{u - 1}{u ^{4}} : \frac{1}{u ^{3}} - \frac{1}{u ^{4}}

= \int \frac{1}{u ^{3}} - \frac{1}{u ^{4}} d u

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{1}{u ^{3}} d u - \int \frac{1}{u ^{4}} d u

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

\int \frac{1}{u ^{4}} d u = - \frac{1}{3 u ^{3}}

= - \frac{1}{2 u ^{2}} - (- \frac{1}{3 u ^{3}})

u = t + 1

代回

= - \frac{1}{2 ( t + 1 ) ^{2}} - (- \frac{1}{3 ( t + 1 ) ^{3}})

化简

= - \frac{1}{2 ( t + 1 ) ^{2}} + \frac{1}{3 ( t + 1 ) ^{3}}

解答补常数

= - \frac{1}{2 ( t + 1 ) ^{2}} + \frac{1}{3 ( t + 1 ) ^{3}} + C

作图

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