Question

How do you integrate `(9x)/(9x^2+3x-2)` using partial fractions?

Answer 1

The answer is `==1/3ln(3x-1)+2/3ln(3x+2)+C`

Explanation:

Let's factorise the denominator

`9x^2+3x-2=(3x-1)(3x+2)`

Now, we can start by the decomposition into partial fractions

`(9x)/(9x^2+3x-2)=(9x)/((3x-1)(3x+2))=A/(3x-1)+B/(3x+2)`

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

Therefore,

`9x=A(3x+2)+B(3x-1)`

Let `x=0`, `=>`, `0=2A-B`

Coefficients of `x`,

`9=3A+3B`, `=>`, `A+B=3`

Solving for A and B, we get `A=1` and `B=2`

So,
`(9x)/(9x^2+3x-2)=1/(3x-1)+2/(3x+2)`

`int(9xdx)/(9x^2+3x-2)=intdx/(3x-1)+int(2dx)/(3x+2)`

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