integral from 0 to 5 of (5t)/((t-6)^2)

解答

\int_{0}^{5} \frac{5 t}{( t - 6 ) ^{2}} d t

5 (- ln (6) + 5)

十进制

16.04120 \dots

求解步骤

\int_{0}^{5} \frac{5 t}{( t - 6 ) ^{2}} d t

提出常数:

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

= 5 \cdot \int_{0}^{5} \frac{t}{( t - 6 ) ^{2}} d t

使用换元积分法

= 5 \cdot \int_{- 6}^{- 1} \frac{u + 6}{u ^{2}} d u

乘开

\frac{u + 6}{u ^{2}} : \frac{1}{u} + \frac{6}{u ^{2}}

= 5 \cdot \int_{- 6}^{- 1} \frac{1}{u} + \frac{6}{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

= 5 (\int_{- 6}^{- 1} \frac{1}{u} d u + \int_{- 6}^{- 1} \frac{6}{u ^{2}} d u)

\int_{- 6}^{- 1} \frac{1}{u} d u = - ln (6)

\int_{- 6}^{- 1} \frac{6}{u ^{2}} d u = 5

= 5 (- ln (6) + 5)

x \to 1 lim (2 x)

\frac{d}{d x} (a x^{- 2})

d er i v a t i v e y = sin (cos (tan (π x)))

\frac{d}{d x} (0.006 x^{3} + 0.07 x^{2} + 18 x + 600)

\frac{d y}{d x} = y (2 x - 5)