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

解答

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

\frac{1}{3} ln ∣ x + 2 ∣ + \frac{2}{3} ln ∣ x - 4 ∣ + C

求解步骤

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

将

\frac{x}{( x + 2 ) ( x - 4 )}

用部份分式展开:

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

= \int \frac{1}{3 ( x + 2 )} + \frac{2}{3 ( x - 4 )} d x

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{1}{3 ( x + 2 )} d x + \int \frac{2}{3 ( x - 4 )} d x

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

\int \frac{2}{3 ( x - 4 )} d x = \frac{2}{3} ln ∣ x - 4 ∣

= \frac{1}{3} ln ∣ x + 2 ∣ + \frac{2}{3} ln ∣ x - 4 ∣

解答补常数

= \frac{1}{3} ln ∣ x + 2 ∣ + \frac{2}{3} ln ∣ x - 4 ∣ + C

\int sec (2 x) \cdot tan (2 x) d x

\int \frac{2 x - 3}{x ^{2} + 4} d x

n = 1 \sum \infty \frac{2 n}{n ^{2}}

d er i v a t i v e y = - 8 x ln (6 x + 5)

t an g e n t y = (x^{3} - 16 x)^{12}, (4, 0)