Answer
$\dfrac{x+5}{x-1} \quad \text{where } x\ne1$
Work Step by Step
By factoring both denominator and numerator and cancling the common factors, we have
\begin{align*}
\require{cancel}
\dfrac{x^2+4x-5}{x^2-2x+1}&=\dfrac{(x+5)(x-1)}{(x-1)^2}\\
\\&=\frac{(x+5)\cancel{(x-1)}}{(x-1)\cancel{^2}}\\
\\&= \frac{x+5}{x-1}, \quad x\ne1
\end{align*}
