What is the integral of (6x+8)/(x(x+2)^2) ?

The integral of (6x+8)/(x(x+2)^2) is -2/(x+2)+2ln|x|-2ln|x+2|+C

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integral of (6x+8)/(x(x+2)^2)

\int \frac{6 x + 8}{x ( x + 2 ) ^{2}} d x

- \frac{2}{x + 2} + 2 ln ∣ x ∣ - 2 ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{6 x + 8}{x ( x + 2 ) ^{2}} d x

Expand

\frac{6 x + 8}{x ( x + 2 ) ^{2}} : \frac{6 x}{x ( x + 2 ) ^{2}} + \frac{8}{x ( x + 2 ) ^{2}}

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{6 x}{x ( x + 2 ) ^{2}} d x + \int \frac{8}{x ( x + 2 ) ^{2}} d x

\int \frac{6 x}{x ( x + 2 ) ^{2}} d x = - \frac{6}{x + 2}

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

= - \frac{6}{x + 2} + 8 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{2 ( x + 2 )})

Simplify

- \frac{6}{x + 2} + 8 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{2 ( x + 2 )}) : - \frac{2}{x + 2} + 2 ln ∣ x ∣ - 2 ln ∣ x + 2 ∣

= - \frac{2}{x + 2} + 2 ln ∣ x ∣ - 2 ln ∣ x + 2 ∣

Add a constant to the solution

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

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What is the integral of (6x+8)/(x(x+2)^2) ?
The integral of (6x+8)/(x(x+2)^2) is -2/(x+2)+2ln|x|-2ln|x+2|+C