How can I solve the following eq. through partial fraction?
$$\frac{g}{(g-1)(g+1)(bg^2+1)}$$
following is my solution
$$\frac{g}{(g-1)(g+1)(bg^2+1)} = \frac{A}{g-1} + \frac{B}{g+1} + \frac{Cg+D}{bg^2+1}$$.
After solving these equations, I get following 4 equations
$$A+B+C = 0 \ \ \ (1)$$ $$A-B+D = 0 \ \ \ (2)$$ $$A+B-C = 1 \ \ \ (3)$$ $$A-B-B = 0 \ \ \ (4)$$
However, after solving (1) and (4) I get C = 1/2. Which does not seem right when I insert C back into eq. 1 and 3 and both come with a different answer for A+B
Any help will be highly appreciated.
The four equations should be$$\left\{\begin{array}{l}A-B-D=0\\A+B-C=1\\bA-bB+D=0\\bA+bB+C=0,\end{array}\right.$$whose solution is $A=\frac1{2b+2}$, $B=\frac1{2b+2}$, $C=-\frac b{b+1}$, and $D=0$.