What is the integral of (x-5)/((x+4)(x+3)^2) ?

The integral of (x-5)/((x+4)(x+3)^2) is -9ln|x+4|+9ln|x+3|+8/(x+3)+C

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

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

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

Solution steps

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

Take the partial fraction of

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

Apply the Sum Rule:

\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})

Simplify

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

Add a constant to the solution

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

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

x^{2} cos (2 x)

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}))

What is the integral of (x-5)/((x+4)(x+3)^2) ?
The integral of (x-5)/((x+4)(x+3)^2) is -9ln|x+4|+9ln|x+3|+8/(x+3)+C