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

The integral of (x-4)/((x+4)(x+3)^2) is -8ln|x+4|+8ln|x+3|+7/(x+3)+C

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

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

- 8 ln ∣ x + 4 ∣ + 8 ln ∣ x + 3 ∣ + \frac{7}{x + 3} + C

Solution steps

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

Take the partial fraction of

\frac{x - 4}{( x + 4 ) ( x + 3 ) ^{2}} : - \frac{8}{x + 4} + \frac{8}{x + 3} - \frac{7}{( x + 3 ) ^{2}}

= \int - \frac{8}{x + 4} + \frac{8}{x + 3} - \frac{7}{( 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{8}{x + 4} d x + \int \frac{8}{x + 3} d x - \int \frac{7}{( x + 3 ) ^{2}} d x

\int \frac{8}{x + 4} d x = 8 ln ∣ x + 4 ∣

\int \frac{8}{x + 3} d x = 8 ln ∣ x + 3 ∣

\int \frac{7}{( x + 3 ) ^{2}} d x = - \frac{7}{x + 3}

= - 8 ln ∣ x + 4 ∣ + 8 ln ∣ x + 3 ∣ - (- \frac{7}{x + 3})

Simplify

= - 8 ln ∣ x + 4 ∣ + 8 ln ∣ x + 3 ∣ + \frac{7}{x + 3}

Add a constant to the solution

= - 8 ln ∣ x + 4 ∣ + 8 ln ∣ x + 3 ∣ + \frac{7}{x + 3} + C

\frac{d}{d x} (\frac{1}{5} x^{5} - \frac{1}{2 x ^{4}} + 3 x + 9)

y^{^{''}} + 5 y^{^{'}} - 14 y = 0

\frac{d}{d x} (\frac{x}{2 e ^{x}})

\frac{\partial}{\partial x} (15 x^{2} - 25 x y + 9 y^{2})

x \to 8 lim (5 x + ∣ x - 8 ∣)

What is the integral of (x-4)/((x+4)(x+3)^2) ?
The integral of (x-4)/((x+4)(x+3)^2) is -8ln|x+4|+8ln|x+3|+7/(x+3)+C