Answer
$\frac{2}{x-1}+\frac{4}{(x-1)^{2}}$
Work Step by Step
$\frac{2x+2}{x^{2}-2x+1}=\frac{2x+2}{(x-1)^{2}}$
Decomposing to partial fractions, we get
$\frac{2x+2}{(x-1)^{2}}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}$
This gives A(x-1)+B= 2x+2.
Equating the coefficients of x and the constant term, we get:
A=2, -A+B=2 or B=4
Thus we can write
$\frac{2x+2}{x^{2}-2x+1}=\frac{2}{x-1}+\frac{4}{(x-1)^{2}}$
