Question

How do you integrate `int (4x - 16) / (x^2 - 2x - 3)` using partial fractions?

How do you integrate `int (4x - 16) / (x^2 - 2x - 3)dx` using partial fractions?

Answer 1

`-ln|x-3|+5ln|x+1|+C`,

OR,

`ln|(x+1)^5/(x-3)|+C`.

Explanation:

Let `I=int(4x-16)/(x^2-2x-3)dx=4int(x-4)/((x-3)(x+1))dx`

We decompose the Integrand

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

Using Heavyside's Cover-up Method to find consts. `A,B in RR`,

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

`B=[(x-4)/(x-3)]_(x=-1)=(-1-4)/(-1-3)=5/4`.

Hence, `I=4int[(-1/4)/(x-3)+(5/4)/(x+1)]dx`

`=-int1/(x-3)dx+5int1/(x+1)dx`

`:. I=-ln|x-3|+5ln|x+1|+C`, or,

`I=ln|(x+1)^5/(x-3)|+C`.

Enjoy Maths.!