integral from 0 to 7 of (3t)/((t-8)^2)

解答

\int_{0}^{7} \frac{3 t}{( t - 8 ) ^{2}} d t

3 (- 3 ln (2) + 7)

十进制

14.76167 \dots

求解步骤

\int_{0}^{7} \frac{3 t}{( t - 8 ) ^{2}} d t

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int_{0}^{7} \frac{t}{( t - 8 ) ^{2}} d t

使用换元积分法

= 3 \cdot \int_{- 8}^{- 1} \frac{u + 8}{u ^{2}} d u

乘开

\frac{u + 8}{u ^{2}} : \frac{1}{u} + \frac{8}{u ^{2}}

= 3 \cdot \int_{- 8}^{- 1} \frac{1}{u} + \frac{8}{u ^{2}} 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

= 3 (\int_{- 8}^{- 1} \frac{1}{u} d u + \int_{- 8}^{- 1} \frac{8}{u ^{2}} d u)

\int_{- 8}^{- 1} \frac{1}{u} d u = - 3 ln (2)

\int_{- 8}^{- 1} \frac{8}{u ^{2}} d u = 7

= 3 (- 3 ln (2) + 7)

t an g e n t f (x) = \frac{2}{3 x + 5}, at a = - 1

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ma c l a u r in x sin (- 2 x) π

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