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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Take the partial fraction of

\frac{1}{u ^{2} ( u - 2 )} : - \frac{1}{4 u} - \frac{1}{2 u ^{2}} + \frac{1}{4 ( u - 2 )}

= \int - \frac{1}{4 u} - \frac{1}{2 u ^{2}} + \frac{1}{4 ( u - 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}{4 u} d u - \int \frac{1}{2 u ^{2}} d u + \int \frac{1}{4 ( u - 2 )} d u

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

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

\int \frac{1}{4 ( u - 2 )} d u = \frac{1}{4} ln ∣ u - 2 ∣

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

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