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Section 7.4 : Partial Fractions
5. Evaluate the integral \( \displaystyle \int_{2}^{4}{{\frac{{3{z^2} + 1}}{{\left( {z + 1} \right){{\left( {z - 5} \right)}^2}}}\,dz}}\).
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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.
\[\frac{{3{z^2} + 1}}{{\left( {z + 1} \right){{\left( {z - 5} \right)}^2}}} = \frac{A}{{z + 1}} + \frac{B}{{z - 5}} + \frac{C}{{{{\left( {z - 5} \right)}^2}}}\] Show Step 2Setting 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 3We 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 4We can now do the integral.
\[\begin{align*}\int_{2}^{4}{{\frac{{3{z^2} + 1}}{{\left( {z + 1} \right){{\left( {z - 5} \right)}^2}}}\,dz}} & = \int_{2}^{4}{{\frac{{\frac{1}{9}}}{{z + 1}} + \frac{{\frac{{26}}{9}}}{{z - 5}} + \frac{{\frac{{38}}{3}}}{{{{\left( {z - 5} \right)}^2}}}\,dz}}\\ & = \left. {\left( {\frac{1}{9}\ln \left| {z + 1} \right| + \frac{{26}}{9}\ln \left| {z - 5} \right| - \frac{{\frac{{38}}{3}}}{{z - 5}}} \right)} \right|_2^4 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{9}\ln \left( 5 \right) - \frac{{27}}{9}\ln \left( 3 \right) + \frac{{76}}{9}}}\end{align*}\]