What is the integral of (4x+6)/((x+1)^2) ?

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

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integral of (4x+6)/((x+1)^2)

\int \frac{4 x + 6}{( x + 1 ) ^{2}} d x

4 ln ∣ x + 1 ∣ - \frac{2}{x + 1} + C

Solution steps

\int \frac{4 x + 6}{( x + 1 ) ^{2}} d x

Expand

\frac{4 x + 6}{( x + 1 ) ^{2}} : \frac{4 x}{( x + 1 ) ^{2}} + \frac{6}{( 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{4 x}{( x + 1 ) ^{2}} d x + \int \frac{6}{( x + 1 ) ^{2}} d x

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

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

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

Simplify

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

= 4 ln ∣ x + 1 ∣ - \frac{2}{x + 1}

Add a constant to the solution

= 4 ln ∣ x + 1 ∣ - \frac{2}{x + 1} + C

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