# compute the indefinite integral int 1/((x-1)(x+1)^2) using partial fraction

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### 1 Answer

You should decompose the fraction in simpler partial fractions, hence you will solve several less sophisticated integrals instead solving the integral of original fraction.

`1/((x-1)(x+1)^2) = A/(x - 1) + B/(x + 1) + C/(x + 1)^2`

You need to bring the fractions to a common denominator such that:

`1 = A(x + 1)^2 + B (x - 1)(x + 1) + C(x - 1)`

Opening the brackets yields:

`1 = Ax^2 + 2Ax + A + Bx^2 - B + Cx - C`

You need to factor out like powers of x:

`1 = x^2(A+B) + x(2A + C) + A - B - C`

Equating the coefficients of like powers yields:

A+B = 0 => A = -B

2A+C = 0 => C = -2A

`A - B - C = 1 =gt A + A + 2A = 1 =gt 4A = 1 =gt A = 1/4`

`B = -1/4; C = -1/2`

`1/((x-1)(x+1)^2) = 1/(4(x - 1))- 1/(4(x + 1))- 1/(2(x + 1)^2)`

`` `int (dx)/((x-1)(x+1)^2) = int (dx)/(4(x - 1))- int (dx)/(4(x + 1))- int (dx)/(2(x + 1)^2)`

`int (dx)/((x-1)(x+1)^2) = (1/4)ln|x - 1| - (1/4)ln|x+1| - (1/2)int (dx)/((x + 1)^2)`

You should come up with the substitution `x + 1 = y =gt dx = dy` , hence `int (dx)/((x + 1)^2) = int dy/y^2 = -1/y + c.`

`(1/2)int (dx)/((x + 1)^2) = -1/(2(x + 1)) + c`

`int (dx)/((x-1)(x+1)^2) = (1/4)ln|x - 1| - (1/4)ln|x+1| -1/(2(x + 1)) + c`

`int (dx)/((x-1)(x+1)^2) = (1/4) ln|(x-1)/(x+1)| - 1/(2(x + 1)) + c`

**Hence, evaluating the indefinite integral yields **

**`int (dx)/((x-1)(x+1)^2) = (1/4) ln|(x-1)/(x+1)| - 1/(2(x + 1)) + c.` **