How do you integrate $\frac{x^{3} + 1}{{(x^{2} - 1)}^{2} (x + 1)}$ $(x^3+1)/((x^2-1)^2(x+1))$ using partial fractions?

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Answer 1 · 2016-09-25T01:54:40

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

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

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:

$1³ + 1 = E(1 + 1)³$
$2 = E(2)³$
$E = 1/4$

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

$A = 0, B = 3/4 and D = 0$

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.

How do you integrate (x^3+1)/((x^2-1)^2(x+1))x3+1(x2−1)2(x+1)(x^3+1)/((x^2-1)^2(x+1)) using partial fractions?