how do i integrate this
$\displaystyle \frac{1}{(3x^2-1)(1-x^2)} $
I have to split this into partial fraction but I don't if this is correct
$\displaystyle \frac{1}{(3x^2-1)(1-x^2)} =\frac{A}{(1-x^2)} + \frac{B}{3x^2-1}$
how do i integrate this
$\displaystyle \frac{1}{(3x^2-1)(1-x^2)} $
I have to split this into partial fraction but I don't if this is correct
$\displaystyle \frac{1}{(3x^2-1)(1-x^2)} =\frac{A}{(1-x^2)} + \frac{B}{3x^2-1}$
Note that $\displaystyle 1-x^2 = (1+x)(1-x) $, and that, similarly, $\displaystyle 3x^2 - 1 $ can be written as $\displaystyle (\sqrt{3}x)^2 -1 $, which can be written $\displaystyle (\sqrt{3}x - 1)(\sqrt{3}x+1) $
Knowing this, you can write:
$\displaystyle \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 \displaystyle \int \frac{dx}{ax+b} = \frac{1}{a} \ln(ax+b) + C $