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?

you are mistaken in the second step, the expression must be simplified as 
$$2x^2-1=A(x-1)(2x^2+3)+B(x+1)(2x^2+3)+(C+Dx)(x^2-1)$$
now, compare powers of $x^3, x^2, x$ & constant terms, you will get
$$2A+2B+D=0\tag 1$$ 
 $$-2A+2B+C=2\tag 2$$
$$3A+3B-D=0\tag 3$$
$$-3A+3B-C=-1\tag 4$$
solve all four linear equations to get the values of A, B,C & D you will get
$$A=-\frac{1}{10}, B=\frac{1}{10}, C=\frac{8}{5}, D=0$$
