Question

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

Answer 1

Let's first determine a factorisation for `x^3 + x^2 - x- 1`.

`x^3 + x^2 - x - 1 = x^2(x + 1) -(x + 1) = (x^2 - 1)(x + 1) = (x + 1)^2(x - 1)`

The partial fraction decomposition will be of the form

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

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

`A(x^2 + 2x + 1) + Bx^2 + Cx - Bx - C = 3x - 2`

`Ax^2 + 2Ax + A + Bx^2 + Cx - Bx -C = 3x - 2`

`(A + B)x^2 + (2A - B + C)x + (A - C) = 3x - 2`

We now get the system of equations `{(A + B = 0), (2A - B + C = 3), (A - C =-2):}`

Solve to get

`2A -(-A) + A +2 = 3`

`A = 1/4`

`B = -1/4`

`C = 9/4`

The integral becomes

`int 1/(4(x - 1)) - (1/4x -9/4)/(x + 1)^2 dx`

`int 5/(4(x - 1))dx - int (x - 9)/(4(x + 1)^2)dx`

We can rewrite `x - 9` as `(x + 1)-10`. Therefore, the second integral is equivalent to

`-1/4(int1/(x + 1) - 10/(x + 1)^2)dx`

This can be integrated as

`-1/4ln|x + 1| - 5/(2(x + 1))`

The other part of the integral is equal to `1/4ln|x - 1|`. Putting all of this together, we have

`1/4ln|x - 1| - 1/4ln|x + 1| - 5/(2(x + 1))+ C`

By laws of logarithms, we have

`1/4ln|(x - 1)/(x +1)| -5/(2(x + 1))+ C`

Hopefully this helps!