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

Feb 17, 2016

$\frac{3}{4} \ln | x - 3 | + \frac{1}{4} \ln | x + 1 | + c$

#### Explanation:

begin by factorising the denominator

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

$\Rightarrow \frac{x}{\left(x - 3\right) \left(x + 1\right)} = \frac{A}{x - 3} + \frac{B}{x + 1}$

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 $\Rightarrow B = \frac{1}{4}$

let x = 3 in (1) : 3 = 4A $\Rightarrow A = \frac{3}{4}$

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

$= \frac{3}{4} \ln | x - 3 | + \frac{1}{4} \ln | x + 1 | + c$