Setting the numerators equal gives,

\[3{z^2} + 1 = A{\left( {z - 5} \right)^2} + B\left( {z + 1} \right)\left( {z - 5} \right) + C\left( {z + 1} \right)\] Show Step 3

We can use the “trick” discussed in the notes to easily get two of the coefficients and then we can just pick another value of \(z\) to get the third so let’s do that. Here is that work.

\[\begin{align*}z = & - 1: & 4 = & \,36A\\ & & & \\ z = & \,5 \,\,\,\,\,\, : & 76 = & \,6C\\ & & & \\z = & \,0 \,\,\,\,\,\, : & 1 = & \,25A - 5B + C = \frac{{139}}{9} - 5B\end{align*}\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\begin{aligned} & {A = \frac{1}{9}}\\& {B = \frac{{26}}{9}}\\ & {C = \frac{{38}}{3}}\end{aligned}\]

The partial fraction form of the integrand is then,

\[\frac{{3{z^2} + 1}}{{\left( {z + 1} \right){{\left( {z - 5} \right)}^2}}} = \frac{{\frac{1}{9}}}{{z + 1}} + \frac{{\frac{{26}}{9}}}{{z - 5}} + \frac{{\frac{{38}}{3}}}{{{{\left( {z - 5} \right)}^2}}}\] Show Step 4

We can now do the integral.

\[\begin{align*}\int_{2}^{4}{{\frac{{3{z^2} + 1}}{{\left( {z + 1} \right){{\left( {z - 5} \right)}^2}}}\,dz}} & = \int_{2}^{4}{{\frac{{\frac{1}{9}}}{{z + 1}} + \frac{{\frac{{26}}{9}}}{{z - 5}} + \frac{{\frac{{38}}{3}}}{{{{\left( {z - 5} \right)}^2}}}\,dz}}\\ & = \left. {\left( {\frac{1}{9}\ln \left| {z + 1} \right| + \frac{{26}}{9}\ln \left| {z - 5} \right| - \frac{{\frac{{38}}{3}}}{{z - 5}}} \right)} \right|_2^4 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{9}\ln \left( 5 \right) - \frac{{27}}{9}\ln \left( 3 \right) + \frac{{76}}{9}}}\end{align*}\]