How do you integrate $int x^3/[x(x^2+2x+1)]$ using partial fractions?

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Answer 1 · 2018-04-05T15:47:02

$x-2ln|x+1|-1/(x+1)+C$

$x^3/[x(x^2+2x+1)] = x^2/(x+1)^2 =(1-1/(x+1))^2$
$qquad = 1-2/(x+1)+1/(x+1)^2$

Thus

$int x^3/[x(x^2+2x+1)]dx = int ( 1-2/(x+1)+1/(x+1)^2)dx$
$qquad = x-2ln|x+1|-1/(x+1)+C$

How do you integrate int x^3/[x(x^2+2x+1)]int x^3/[x(x^2+2x+1)] using partial fractions?