integral of (x+2)/(x-2)(x^2-4x)

解答

\int \frac{x + 2}{x - 2} (x^{2} - 4 x) d x

2 x^{2} - 12 x + \frac{1}{3} (x - 2)^{3} - 16 ln ∣ x - 2 ∣ + 16 + C

求解步骤

\int \frac{x + 2}{x - 2} (x^{2} - 4 x) d x

\frac{x + 2}{x - 2} = \frac{4}{x - 2} + 1

= \int (\frac{4}{x - 2} + 1) (x^{2} - 4 x) d x

乘开

(\frac{4}{x - 2} + 1) (x^{2} - 4 x) : \frac{x ^{3} - 2 x ^{2} - 8 x}{x - 2}

= \int \frac{x ^{3} - 2 x ^{2} - 8 x}{x - 2} d x

乘开

\frac{x ^{3} - 2 x ^{2} - 8 x}{x - 2} : \frac{x ^{3}}{x - 2} - \frac{2 x ^{2}}{x - 2} - \frac{8 x}{x - 2}

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{x ^{3}}{x - 2} d x - \int \frac{2 x ^{2}}{x - 2} d x - \int \frac{8 x}{x - 2} d x

\int \frac{x ^{3}}{x - 2} d x = 3 x^{2} + 8 ln ∣ x - 2 ∣ + \frac{1}{3} (x - 2)^{3} - 12

\int \frac{2 x ^{2}}{x - 2} d x = x^{2} + 4 x + 8 ln ∣ x - 2 ∣ - 12

\int \frac{8 x}{x - 2} d x = 8 (x - 2 + 2 ln ∣ x - 2 ∣)

= 3 x^{2} + 8 ln ∣ x - 2 ∣ + \frac{1}{3} (x - 2)^{3} - 12 - (x^{2} + 4 x + 8 ln ∣ x - 2 ∣ - 12) - 8 (x - 2 + 2 ln ∣ x - 2 ∣)

化简

3 x^{2} + 8 ln ∣ x - 2 ∣ + \frac{1}{3} (x - 2)^{3} - 12 - (x^{2} + 4 x + 8 ln ∣ x - 2 ∣ - 12) - 8 (x - 2 + 2 ln ∣ x - 2 ∣) : 2 x^{2} - 12 x + \frac{1}{3} (x - 2)^{3} - 16 ln ∣ x - 2 ∣ + 16

= 2 x^{2} - 12 x + \frac{1}{3} (x - 2)^{3} - 16 ln ∣ x - 2 ∣ + 16

解答补常数

= 2 x^{2} - 12 x + \frac{1}{3} (x - 2)^{3} - 16 ln ∣ x - 2 ∣ + 16 + C

n = 2 \sum \infty \frac{1}{2 n ln ( n )}

\frac{\partial}{\partial x} (x cos (x) y)

\frac{d}{d x} (e^{- x^{8}})

a re a x^{3} - 12 x^{2} + 27 x, - x^{3} + 12 x^{2} - 27 x

d er i v a t i v e x^{\frac{x}{3}}