I need help with the following:
$\frac{1}{(x-1)^2(x+1)}$
Using the "cover-up" method, I can identify the numerator of the fraction with the denominator $(x+1)$ fairly quickly, and it is $1/4$. Then I arrive at the following: $$1 \equiv (1/4)(x-1)^2 + B(x-1)(x+1)+(Dx+E)(x+1)$$
Then I have three equations with two unknowns in each, a total of three unknowns across three equations. Which is not solvable. What do I do next after the step above?
Let $\displaystyle \frac{1}{(x-1)^2\cdot (x+1)} = \frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)}$
$\displaystyle 1 = A(x-1)\cdot (x+1)+B(x+1)+C(x-1)^2...........(1)$
Put $(x-1) = 0\Rightarrow x=1\;,$ in above equation. So we get $\displaystyle 1 = 2B\Rightarrow B = \frac{1}{2}$
Similarly Put $(x+1) = 0\Rightarrow x=-1\;,$ in above equation, So We get $\displaystyle 1 = 4C\Rightarrow C = \frac{1}{4}$
Now Equating Coefficient of constant term in $(1)$ equation.
So Put $x=0$ on both side, We get $\displaystyle 1=-A+B+C\Rightarrow 1=-A+\frac{1}{2}+\frac{1}{4}$
so we get $\displaystyle A = -\frac{1}{4}$
Hint:
You need to match the coefficients of all $x^k$ terms on both sides of the equation. The left side ($=1$) has coefficients equal to $0$ for all terms $x^1$ and larger.
