Question

How do you integrate `int (2x+1)/(4x^2+12x-7) dx` using partial fractions?

Answer 1

The integral `int((2x+1)dx)/(4x^2+12x-7)=3/8ln(2x+7)+1/8ln(2x-1)+C`

Explanation:

Let's factorise the denominator
`4x^2+12x-7=(2x+7)(2x-1)`

We look for the partial fractions
`(2x+1)/(4x^2+12x-7)=A/(2x+7)+B/(2x-1)`

`(2x+1)/(4x^2+12x-7)=(A(2x-1)+B(2x+7))/(4x^2+12x-7)`

so, `2x+1=A(2x-1)+B(2x+7))`
Let `x=1/2` `=>` `2=8B` `=>` `B=1/4`
Let `x=0` `=>` `1=-A+7/4` `=>` `A=3/4`

`(2x+1)/(4x^2+12x-7)=(3/4)/(2x+7)+(1/4)/(2x-1)`
Let's do the integral
`int((2x+1)dx)/(4x^2+12x-7)=int((3/4)dx)/(2x+7)+int((1/4)dx)/(2x-1)`

`=3/8ln(2x+7)+1/8ln(2x-1)+C`