Setting the numerators equal gives,

\[{z^2} + 2z + 3 = A\left( {{z^2} + 4} \right) + \left( {Bz + C} \right)\left( {z - 6} \right) = \left( {A + B} \right){z^2} + \left( { - 6B + C} \right)z + 4A - 6C\]

In this case the “trick” discussed in the notes won’t work all that well for us and so we’ll have to resort to multiplying everything out and collecting like terms as shown above.

Show Step 3

Now, setting the coefficients equal gives the following system.

\[\begin{align*}{{z^2}:}& \hspace{0.25in} & A + B = & \, 1\\ & \\ {{z^1}:}& \hspace{0.25in}& - 6B + C = & \, 2\\ & \\{{z^0}:}& \hspace{0.25in} & 4A - 6C = & \, 3\end{align*}\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\begin{aligned}&{A = \frac{{51}}{{40}}}\\&{B = - \frac{{11}}{{40}}}\\&{C = \frac{7}{{20}}}\end{aligned}\]

The partial fraction form of the integrand is then,

\[\frac{{{z^2} + 2z + 3}}{{\left( {z - 6} \right)\left( {{z^2} + 4} \right)}} = \frac{{\frac{{51}}{{40}}}}{{z - 6}} + \frac{{ - \frac{{11}}{{40}}z + \frac{7}{{20}}}}{{{z^2} + 4}}\] Show Step 4

We can now do the integral.

\[\begin{align*}\int{{\frac{{{z^2} + 2z + 3}}{{\left( {z - 6} \right)\left( {{z^2} + 4} \right)}}\,dz}} & = \int{{\frac{{\frac{{51}}{{40}}}}{{z - 6}} + \frac{{ - \frac{{11}}{{40}}z + \frac{7}{{20}}}}{{{z^2} + 4}}\,dz}}\\ & = \int{{\frac{{\frac{{51}}{{40}}}}{{z - 6}} + \frac{{ - \frac{{11}}{{40}}z}}{{{z^2} + 4}} + \frac{{\frac{7}{{20}}}}{{{z^2} + 4}}\,dz}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{51}}{{40}}\ln \left| {z - 6} \right| - \frac{{11}}{{80}}\ln \left| {{z^2} + 4} \right| + \frac{7}{{40}}{{\tan }^{ - 1}}\left( {\frac{z}{2}} \right) + c}}\end{align*}\]

Note that the second integration needed the substitution \(u = {z^2} + 4\) while the third needed the formula provided in the notes.