how do i integrate this
$\displaystyle \frac{1}{(3x^2-1)^2(1-x^2)} $
I have to split this into partial fraction but I don't know how
$\displaystyle \frac{1}{(3x^2-1)^2(1-x^2)} =\frac{Ax+B}{(1-x^2)} + \frac{Cx+D}{(3x^2-1)^2}?$
how do i integrate this
$\displaystyle \frac{1}{(3x^2-1)^2(1-x^2)} $
I have to split this into partial fraction but I don't know how
$\displaystyle \frac{1}{(3x^2-1)^2(1-x^2)} =\frac{Ax+B}{(1-x^2)} + \frac{Cx+D}{(3x^2-1)^2}?$
Note that $\displaystyle 3x^2-1=\left(\sqrt{3}x-1\right)\left(\sqrt{3}x+1\right)$ and $\displaystyle 1-x^2=\left(1-x\right)\left(1+x\right)$.
So we have $\displaystyle \frac{1}{\left[(\sqrt{3}x+1)(\sqrt{3}x-1)\right]^2(1+x)(1-x)}=\frac{A}{\sqrt{3}x-1}+\frac{B}{(\sqrt{3}x-1)^2}+\frac{C}{\sqrt{3}x+1}$ $\displaystyle +\frac{D}{(\sqrt{3}x+1)^2}+\frac{E}{1+x}+\frac{F}{ 1-x}$
This is going to get quite messy...feel free to post back when you get stuck!