# How do you integrate (x^3+1)/((x^2-1)^2(x+1)) using partial fractions?

Sep 25, 2016

Rewrite the denominator as (x + 1)³(x - 1)² and then proceed with the normal expansion process

#### Explanation:

(x³ + 1)/((x + 1)³(x - 1)²) =
A/((x + 1)) + B/((x + 1)²) + C/((x + 1)³) + D/(x - 1) + E/((x - 1)²

Multiply both sides by the denominator:

x³ + 1 = A(x + 1)²(x - 1)² + B(x + 1)(x - 1)² + C(x - 1)² + D(x + 1)³(x - 1) + E(x + 1)³

Let x = -1:

-1³ + 1 = C(-1 - 1)²
0 = C(-2)²
$C = 0$

Let x = 1:

1³ + 1 = E(1 + 1)³
2 = E(2)³
$E = \frac{1}{4}$

If you continue by creating 3 equations to solve for the remaining 3 unknow values, you will find that:

$A = 0 , B = \frac{3}{4} \mathmr{and} D = 0$

This gives you two integrals:

3/4int 1/((x + 1)²)dx + 1/4int 1/((x - 1)²)dx

Both integrals are trivial and found in lists of integrals.