Section 7.4 : Partial Fractions
7. Evaluate the integral \( \displaystyle \int{{\frac{{{z^2} + 2z + 3}}{{\left( {z - 6} \right)\left( {{z^2} + 4} \right)}}\,dz}}\).
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Start SolutionIn this case the denominator is already factored and so we can go straight to the form of the partial fraction decomposition for the integrand.
\[\frac{{{z^2} + 2z + 3}}{{\left( {z - 6} \right)\left( {{z^2} + 4} \right)}} = \frac{A}{{z - 6}} + \frac{{Bz + C}}{{{z^2} + 4}}\] Show Step 2Setting 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 3Now, 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 4We 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.