Question

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

Answer 1

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

Explanation:

`x^3 - x^2` can be factored as `x^2(x - 1)`.

`(Ax + B)/(x^2) + C/(x - 1) = 2/((x^2)(x - 1))`

`(Ax + B)(x- 1) + C(x^2) = 2`

`Ax^2 + Bx - Ax - B + Cx^2 = 2`

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

We can now write a system of equations.

`{(A + C = 0), (B - A = 0), (-B = 2):}`

Solving, we obtain `B = -2`, `A = -2`, `C = 2`.

Therefore, the partial fraction decomposition is:

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

We now decompose `(-2x - 2)/x^2` into partial fractions.

`A/x^2 + B/x = (-2x- 2)/x^2`

`A + Bx = -2x - 2 -> A = -2 and B = -2`

Therefore, the complete partial fraction decomposition of `2/(x^3 - x^2)` is `-2/x^2 - 2/x + 2/(x - 1)`.

This can be integrated as follows:

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

`=-2ln|x| + 2ln|x- 1| - int(2x^-2) + C`

`=2ln|x - 1| - 2ln|x| -(-2x^-1) + C`

`=2ln|x- 1| - 2ln|x| + 2/x + C`

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

Hopefully this helps!