Answer
Yes, see explanations.
Work Step by Step
Step 1. Find a common denominator and add up the rational polynomials to get:
$\frac{2(x-1)(x+1)+(x+1)+(x-1)^2}{(x+1)(x-1)^2}=\frac{3x^2-x}{(x+1)(x-1)^2}$
Step 2. To decompose the above rational polynomial, assume it will take the form:
$\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{(x-1)^2}$
Step 3. Use a common denominator, add up the terms above to get:
$\frac{A(x-1)^2+B(x+1)(x-1)+C(x+1)}{(x+1)(x-1)^2}=\frac{(A+B)x^2+(-2A+C)x+(A-B+C)}{(x+1)(x-1)^2}$
Step 4. Compare the above results with the expression of setp-1 to set up the following equations:
\begin{cases} A+B=3 \\ -2A+C=-1\\A-B+C=0 \end{cases}
Step 5. Use $B=3-A, C=2A-1$ for substitution, we get $A=1, B=2, C=1$
Step 6. The decomposed form in step-2 becomes:
$\frac{1}{x+1}+\frac{2}{x-1}+\frac{1}{(x-1)^2}$
which is the same as the original expression given in the Exercise.
