find partial fraction $(2x^2-1)/((x^2-1)(2 x^2+3))$ 
Find partial fraction of $\cfrac{(2x^2-1)}{(x^2-1)(2 x^2+3)}$?


My attempt:
I did google and I tried to solved it as :
Let’s first get the general form of the partial fraction decomposition.
$\cfrac{(2x^2-1)}{(x^2-1)(2 x^2+3)}=\cfrac{A}{(x+1)}+\cfrac{B}{(x-1)}+\cfrac{C+Dx}{(2x^2+3)}$
Setting numerators gives,
$(2x^2-1)=A(x-1)(2x^2+3)+B(x+1)(2x^2+3)+C(x^2-1)$
I stuck here, how to proceed next? 
Can you explain it, please?

Multiply out and compare coefficients. You'll find for example that the coefficient of $x^3$ on the right hand side is $2A+2B$.  As in the left hand side there are zero $x^3$ you conclude $2A+2B=0$.
