integral from 0 to 6 of (3t)/((t-7)^2)

解答

\int_{0}^{6} \frac{3 t}{( t - 7 ) ^{2}} d t

3 (- ln (7) + 6)

十进制

12.16226 \dots

求解步骤

\int_{0}^{6} \frac{3 t}{( t - 7 ) ^{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}^{6} \frac{t}{( t - 7 ) ^{2}} d t

使用换元积分法

= 3 \cdot \int_{- 7}^{- 1} \frac{u + 7}{u ^{2}} d u

乘开

\frac{u + 7}{u ^{2}} : \frac{1}{u} + \frac{7}{u ^{2}}

= 3 \cdot \int_{- 7}^{- 1} \frac{1}{u} + \frac{7}{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_{- 7}^{- 1} \frac{1}{u} d u + \int_{- 7}^{- 1} \frac{7}{u ^{2}} d u)

\int_{- 7}^{- 1} \frac{1}{u} d u = - ln (7)

\int_{- 7}^{- 1} \frac{7}{u ^{2}} d u = 6

= 3 (- ln (7) + 6)

\frac{d}{d x} (\frac{4 x ^{2} + 7 x ^{5}}{x})

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