integral of (x-5)/((x+4)(x+3)^2)

解答

\int \frac{x - 5}{( x + 4 ) ( x + 3 ) ^{2}} d x

- 9 ln ∣ x + 4 ∣ + 9 ln ∣ x + 3 ∣ + \frac{8}{x + 3} + C

求解步骤

\int \frac{x - 5}{( x + 4 ) ( x + 3 ) ^{2}} d x

将

\frac{x - 5}{( x + 4 ) ( x + 3 ) ^{2}}

用部份分式展开:

- \frac{9}{x + 4} + \frac{9}{x + 3} - \frac{8}{( x + 3 ) ^{2}}

= \int - \frac{9}{x + 4} + \frac{9}{x + 3} - \frac{8}{( x + 3 ) ^{2}} 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{9}{x + 4} d x + \int \frac{9}{x + 3} d x - \int \frac{8}{( x + 3 ) ^{2}} d x

\int \frac{9}{x + 4} d x = 9 ln ∣ x + 4 ∣

\int \frac{9}{x + 3} d x = 9 ln ∣ x + 3 ∣

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

= - 9 ln ∣ x + 4 ∣ + 9 ln ∣ x + 3 ∣ - (- \frac{8}{x + 3})

化简

= - 9 ln ∣ x + 4 ∣ + 9 ln ∣ x + 3 ∣ + \frac{8}{x + 3}

解答补常数

= - 9 ln ∣ x + 4 ∣ + 9 ln ∣ x + 3 ∣ + \frac{8}{x + 3} + C

\int_{0}^{1} (3 x + 1)^{5} d x

d er i v a t i v e x^{2} cos (2 x)

t an g e n t e^{x} (- 8 - 3 x + 2 x^{2})

\frac{d}{d x} (\frac{x ^{2}}{x ^{2} - 1})

\frac{d}{d x} (lo g_{3} (x^{2}))