integral of (6x)/((x-5)^2)

解答

\int \frac{6 x}{( x - 5 ) ^{2}} d x

6 (ln ∣ x - 5 ∣ - \frac{5}{x - 5}) + C

求解步骤

\int \frac{6 x}{( x - 5 ) ^{2}} d x

提出常数:

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

= 6 \cdot \int \frac{x}{( x - 5 ) ^{2}} d x

使用换元积分法

= 6 \cdot \int \frac{u + 5}{u ^{2}} d u

乘开

\frac{u + 5}{u ^{2}} : \frac{1}{u} + \frac{5}{u ^{2}}

= 6 \cdot \int \frac{1}{u} + \frac{5}{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

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{5}{u ^{2}} d u = - \frac{5}{u}

= 6 (ln ∣ u ∣ - \frac{5}{u})

u = x - 5

代回

= 6 (ln ∣ x - 5 ∣ - \frac{5}{x - 5})

解答补常数

= 6 (ln ∣ x - 5 ∣ - \frac{5}{x - 5}) + C

x \to 2 lim (6 x^{2} + 1)

\int_{- \infty}^{0} e^{9 x} d x

t^{2} (\frac{d y}{d t}) + 4 t y - y^{3} = 0

s l o p e x^{3} + y^{3} - 6 x y = 0, (\frac{4}{3}, \frac{8}{3})

\int_{2}^{5} 4 x^{3} d x