\(\begin{eqnarray*} \dfrac{2x^3-4x^2+11x-1}{x^2-2x+5}&=&\dfrac{Ax+B}{x^2-2x+5}+\dfrac{Cx+D}{(x^2-2x+5)^2}\\ &=&\dfrac{Ax^3+(-2A+B)x^2+(5A-2B+C)x+(5B+D)}{(x^2-2x+5)^2} \end{eqnarray*}\)
So \(A=2,\quad-2A+B=-4,\quad 5A-2B+C=11,\quad 5B+D=-1\).
Since \(A=2\) it follows from the second equation that \(B=0\). Since \(B=0\) it follows from the fourth equation that \(D=-1\). Since \(A=2\) and \(B=0\) it follows from the third equation that \(C=1\). So the decomposition is
\[ \dfrac{2x^3-4x^2+11x-1}{x^2-2x+5}=\dfrac{2x}{x^2-2x+5}+\dfrac{x-1}{(x^2-2x+5)^2} \]