# Thread: Integrate this using partial fraction

1. ## Integrate this using partial fraction

how do i integrate this

$\frac{1}{(3x^2-1)(1-x^2)}$

I have to split this into partial fraction but I don't if this is correct

$\frac{1}{(3x^2-1)(1-x^2)} =\frac{A}{(1-x^2)} + \frac{B}{3x^2-1}$

2. Originally Posted by charikaar
how do i integrate this

$\frac{1}{(3x^2-1)(1-x^2)}$

I have to split this into partial fraction but I don't if this is correct

$\frac{1}{(3x^2-1)(1-x^2)} =\frac{A}{(1-x^2)} + \frac{B}{3x^2-1}$
No.

Factorise the numerator into linear factors and then decompose.

3. Originally Posted by charikaar
how do i integrate this

$\frac{1}{(3x^2-1)(1-x^2)}$

I have to split this into partial fraction but I don't if this is correct

$\frac{1}{(3x^2-1)(1-x^2)} =\frac{A}{(1-x^2)} + \frac{B}{3x^2-1}$
Note that $1-x^2 = (1+x)(1-x)$, and that, similarly, $3x^2 - 1$ can be written as $(\sqrt{3}x)^2 -1$, which can be written $(\sqrt{3}x - 1)(\sqrt{3}x+1)$

Knowing this, you can write:

$\frac{1}{ (\sqrt{3}x - 1)(\sqrt{3}x+1)(1+x)(1-x)} = \frac{A}{\sqrt{3}x - 1}+\frac{B}{\sqrt{3}x + 1} + \frac{C}{1-x} + \frac{D}{1+x}$

When you perform partial fractions, functions like these are extremely easy to integrate using the following rule:

$\displaystyle \int \frac{dx}{ax+b} = \frac{1}{a} \ln(ax+b) + C$