## Precalculus (6th Edition) Blitzer

The partial fraction is, $\frac{6}{\left( x-1 \right)}-\frac{5}{{{\left( x-1 \right)}^{2}}}$
$\frac{6x-11}{{{\left( x-1 \right)}^{2}}}=\frac{A}{\left( x-1 \right)}+\frac{B}{{{\left( x-1 \right)}^{2}}}$ Now, multiply both sides by ${{\left( x-1 \right)}^{2}}$: ${{\left( x-1 \right)}^{2}}\times \frac{6x-11}{{{\left( x-1 \right)}^{2}}}={{\left( x-1 \right)}^{2}}\times \frac{A}{\left( x-1 \right)}+{{\left( x-1 \right)}^{2}}\times \frac{B}{{{\left( x-1 \right)}^{2}}}$ \begin{align} & 6x-11=A\left( x-1 \right)+B \\ & =Ax-A+B \end{align} Then, compare the coefficient of $x$ and constant term: $A=6$ ….. (1) $-A+B=-11$ …… (2) And put the value of A in equation (2): \begin{align} & -6+B=-11 \\ & B=-11+6 \\ & =-5 \end{align} Therefore, $\frac{6x-11}{{{\left( x-1 \right)}^{2}}}=\frac{6}{\left( x-1 \right)}-\frac{5}{{{\left( x-1 \right)}^{2}}}$ Thus, the partial fraction of the provided expression is $\frac{6}{\left( x-1 \right)}-\frac{5}{{{\left( x-1 \right)}^{2}}}$.