Question

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

Answer 1

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

Explanation:

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`