What is the integral of (2x+6)/(x(x+1)^2) ?

The integral of (2x+6)/(x(x+1)^2) is 4/(x+1)+6ln|x|-6ln|x+1|+C

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integral of (2x+6)/(x(x+1)^2)

\int \frac{2 x + 6}{x ( x + 1 ) ^{2}} d x

\frac{4}{x + 1} + 6 ln ∣ x ∣ - 6 ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{2 x + 6}{x ( x + 1 ) ^{2}} d x

Expand

\frac{2 x + 6}{x ( x + 1 ) ^{2}} : \frac{2 x}{x ( x + 1 ) ^{2}} + \frac{6}{x ( x + 1 ) ^{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{2 x}{x ( x + 1 ) ^{2}} d x + \int \frac{6}{x ( x + 1 ) ^{2}} d x

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

\int \frac{6}{x ( x + 1 ) ^{2}} d x = 6 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1})

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

Simplify

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

= \frac{4}{x + 1} + 6 ln ∣ x ∣ - 6 ln ∣ x + 1 ∣

Add a constant to the solution

= \frac{4}{x + 1} + 6 ln ∣ x ∣ - 6 ln ∣ x + 1 ∣ + C

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What is the integral of (2x+6)/(x(x+1)^2) ?
The integral of (2x+6)/(x(x+1)^2) is 4/(x+1)+6ln|x|-6ln|x+1|+C