Question

How do you integrate `(3x+2) / (x^(2)+3x-4)dx` using partial fractions?

Answer 1

`ln|x-1|+2ln|x+4|+C`,

or,

`ln|(x-1)(x+4)^2|+C`.

Explanation:

Let `I=int(3x+2)/(x^2+3x-4)dx=int(3x+2)/((x-1)(x+4))dx`

To decompose the integrand using Method of Partial Fraction, let,

`(3x+2)/((x-1)(x+4))=A/(x-1)+B/(x+4)", where, "A,B in RR`.

We use Heavyside's Cover-up Method to determine `A, and, B` :

`A=[(3x+2)/(x+4)]_(x=1) =(3+2)/(1+4)=1`

`B=[(3x+2)/(x-1)]_(x=-4) =(-12+2)/(-4-1)=2`. Hence,

`I=int[1/(x-1)+2/(x+4)]dx`

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

`=ln|x-1|+2ln|x+4|`

Therefore,

`I=ln|x-1|+2ln|x+4+C|`, or,

`I=ln|(x-1)(x+4)^2|+C`.

Enjoy Maths.!