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?
 A: Well, you have there a polynomial equality, thus you can work taking into account the following: 
(1) Coefficients of corresponding powers of $\;x\;$ in both sides are equal. For example, lets us compare the coefficients of $\;x^3\;$ in both sides:
$$0=2A+2B\implies A=-B$$
and you have a first linear equation for $\;A,B,C,...\;$ 
(2) The equality remains true if you susbtitute $\;x\;$ with any value. For example, let us subsitute in both sides $\;x=1\;$ :
$$1=A\cdot0+B(2)(5)+C\cdot0\implies 10B=1$$
and you've already obtained the value of one of your unknowns.
With the above two operations (perhaps first the second one, and then the first one...) you should be able to work out these problems in most cases.
A: 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 $2A+2B=0$.
A: 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
$$A=-\frac{1}{10}, B=\frac{1}{10}, C=\frac{8}{5}, D=0$$
