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

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

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

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

- \frac{1}{2 ( x ^{2} - 1 )} + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} - \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4 ( x - 1 )} + C

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

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