Answer
$\frac{2}{x-1}-\frac{3}{x+1}+\frac{x}{x^2+2}$
Work Step by Step
Step 1. Factorize the denominator as: $(x^2-1)(x^2+2)=(x-1)(x+1)(x^2+2)$
Step 2. Assume the partial fraction decomposition is:
$\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+2}$
Step 3. Combine the rational functions above:
$\frac{A(x+1)(x^2+2)+B(x-1)(x^2+2)+(Cx+D)(x^1-1)}{(x-1)(x+1)(x^2+2)}=\frac{(A+B+C)x^3+(A-B+D)x^2+(2A+2B-C)x+(2A-2B-D)}{(x-1)(x+1)(x^2+2)}$
Step 4. Compare the above function with the original to setup the following system of equations:
$\begin{cases} A+B+C=0\\A-B+D=5\\2A+2B-C=-3\\2A-2B-D=10 \end{cases}$
Step 5. Solve the above equations: equation1 gives $A+B=-C$, equation3 gives $2(A+B)-C=-3$, combine these two to get $-3C=-3$ or $C=1$. Similarly, equation 2 gives $A-B=5-D$, equation 4 gives $2(A-B)-D=10$, combine these two to get $10-2D-D=10$ or $D=0$. Plug-in the values of $C, D$, we get $A+B=-1, A-B=5$ which gives $A=2, B=-3$
Step 6. The final result is:
$\frac{2}{x-1}-\frac{3}{x+1}+\frac{x}{x^2+2}$
