Answer
$$\frac{2}{{x - 1}} + \frac{4}{{{{\left( {x - 1} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& \frac{{2x + 2}}{{{x^2} - 2x + 1}} \cr
& {\text{Factoring the denominator}} \cr
& \frac{{2x + 2}}{{{{\left( {x - 1} \right)}^2}}} \cr
& \cr
& {\text{The partial fraction decomposition has the form}} \cr
& \frac{{2x + 2}}{{{{\left( {x - 1} \right)}^2}}} = \frac{A}{{x - 1}} + \frac{B}{{{{\left( {x - 1} \right)}^2}}}\,\,\,\left( {\bf{1}} \right) \cr
& \cr
& {\text{Multiplying by }}{\left( {x - 1} \right)^2}{\text{, we have}} \cr
& 2x + 2 = A\left( {x - 1} \right) + B \cr
& 2x + 2 = Ax - A + B \cr
& \cr
& {\text{If we equate coefficients}}{\text{, we get the system}} \cr
& A = 2,\,\,\,\,\,\,\,\,\, - A + B = 2 \to B = 4 \cr
& \cr
& {\text{Then substituting }}A{\text{ and }}B{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& \frac{{2x + 2}}{{{{\left( {x - 1} \right)}^2}}} = \frac{2}{{x - 1}} + \frac{4}{{{{\left( {x - 1} \right)}^2}}} \cr
& {\text{or}} \cr
& \frac{{2x + 2}}{{{x^2} - 2x + 1}} = \frac{2}{{x - 1}} + \frac{4}{{{{\left( {x - 1} \right)}^2}}} \cr} $$
