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

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

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

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

- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

Solution steps

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

Expand

\frac{2 x + 2}{( x + 1 ) ^{2} ( x - 1 )} : \frac{2 x}{( x + 1 ) ^{2} ( x - 1 )} + \frac{2}{( x + 1 ) ^{2} ( x - 1 )}

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

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

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

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

Simplify

- \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{x + 1} + \frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{x + 1} + \frac{1}{2} ln ∣ x - 1 ∣ : - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣

= - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣

Add a constant to the solution

= - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

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