Question

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

Answer 1

`3/4 ln|x-3| + 1/4 ln|x + 1 | + c`

Explanation:

begin by factorising the denominator

`x^2 - 2x - 3 = (x - 3 )(x + 1 )`
Since the factors are linear , the numerators of the partial fractions will be constants , say A and B.

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

multiply through by (x-3)(x+1)

x = A(x+1) + B(x-3) .................................(1)

The aim now is to find values of A and B. Note , that if x = - 1 the term with A will be zero and if x = 3 , the term with B will be zero. This is the starting point to finding A and B.

let x = - 1 in (1) : - 1 = - 4B `rArr B = 1/4`

let x = 3 in (1) : 3 = 4A `rArr A = 3/4`

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

`= 3/4 ln|x-3| + 1/4 ln|x+1| + c`