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

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

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

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

- \frac{1}{2 ( x + 1 )} - \frac{1}{2 ( x - 1 )} + C

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

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

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

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

\int \frac{1}{2 u ^{2}} d u = - \frac{1}{2 u}

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

= - \frac{1}{2 u} - \frac{1}{2 ( u - 2 )}

Substitute back

u = x + 1

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

Simplify

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

Add a constant to the solution

= - \frac{1}{2 ( x + 1 )} - \frac{1}{2 ( x - 1 )} + C

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