Question

How do you find `int (x-1) / ((2x+1) (4x+4)) dx` using partial fractions?

Answer 1

Please see the explanation.

Explanation:

Please notice that the integral can be multiplied by `1/4` and the integrand becomes:

`(x -1)/((2x + 1)(x + 1)) = A/(2x + 1) + B/(x + 1)`

Multiply by the left denominator:

`x -1 = A(x + 1) + B(2x + 1)`

Let `x = -1`

`-2 = -B`

`B = 2`

Let (x = -1/2)

`-3/2 = A(1/2)`

`A = -3`

Check:

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

`(4x +2 - 3x - 3)/((2x + 1)(x + 1)) =`

`(x - 1)/((2x + 1)(x + 1))` This checks:

`int(x -1)/((2x + 1)(4x + 4))dx = 1/2int1/(x + 1)dx - 3/4int 1/(2x + 1)dx`

`int(x -1)/((2x + 1)(4x + 4))dx = 1/2ln|x + 1| - 3/8ln|2x + 1| + C`