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

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Answer 1 · 2017-01-08T00:10:28

Do a partial fraction expansion, check it, then integrate the terms.

Begin with the a partial fraction expansion:

$(x - 1)/(x^3 + x) = A/x + (Bx + C)/(x^2 + 1)$

$x - 1 = A(x^2 + 1) + (Bx + C)x$

$x - 1 = A(x^2 + 1) + Bx^2 + Cx$

To make B and C disappear, let $x = 0$ :

$0 - 1 = A(0^2 + 1)$

A = -1

$x + x^2 = Bx^2 + Cx$

$B = 1$
$C = 1$

Check:

$x/(x^2 + 1)x/x + 1/(x^2 + 1)x/x - 1/x(x^2 + 1)/(x^2 + 1)$

$x^2/(x(x^2 + 1)) + x/(x(x^2 + 1)) - (x^2 + 1)/(x(x^2 + 1))$

$(x -1)/(x(x^2 + 1))$

$(x -1)/(x^3 + x)$

This checks.

$int(x - 1)/(x^3 + x)dx = intx/(x^2 + 1)dx + int1/(x^2 + 1)dx - int1/xdx$

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

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