Section 5.5 : Partial Fractions
5. Determine the partial fraction decomposition of each of the following expression.
\[\frac{{6x + 5}}{{{{\left( {2x - 1} \right)}^2}}}\]Show All Steps Hide All Steps
Start SolutionThe first step is to determine the form of the partial fraction decomposition. For this problem the partial fraction decomposition is,
\[\frac{{6x + 5}}{{{{\left( {2x - 1} \right)}^2}}} = \frac{A}{{2x - 1}} + \frac{B}{{{{\left( {2x - 1} \right)}^2}}}\] Show Step 2The LCD for this expression is \({\left( {2x - 1} \right)^2}\). Adding the terms back up gives,
\[\frac{{6x + 5}}{{{{\left( {2x - 1} \right)}^2}}} = \frac{{A\left( {2x - 1} \right) + B}}{{{{\left( {2x - 1} \right)}^2}}}\] Show Step 3Setting the numerators equal gives,
\[6x + 5 = A\left( {2x - 1} \right) + B\] Show Step 4For this problem we can pick a “good” value of \(x\) to determine only one of the constants. Here is that work.
\[x = \frac{1}{2}: \hspace{0.25in} 8 = B \hspace{0.25in} \to \hspace{0.25in}B = 8\] Show Step 5To get the remaining constant we can use any value of \(x\) and plug that along with the value of \(B\) we found in the previous step into the equation from Step 3.
It really doesn’t matter what value of \(x\) we pick as long as it isn’t \(x = \frac{1}{2}\) since we used that in the previous step. The idea here is to pick a value of \(x\) that won’t create “large” or “messy” numbers, if possible. Good choices are often \(x = 0\) or \(x = 1\), provided they weren’t used in the previous step of course.
For this problem \(x = 0\) seems to be a good choice. Here is the work for this step.
\[\begin{align*}6\left( 0 \right) + 5 & = A\left( {2\left( 0 \right) - 1} \right) + 8\\ 5 & = - A + 8\\ - 3 & = - A\hspace{0.25in} \Rightarrow \hspace{0.25in}A = 3\end{align*}\] Show Step 6The partial fraction decomposition is then,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{6x + 5}}{{{{\left( {2x - 1} \right)}^2}}} = \frac{3}{{2x - 1}} + \frac{8}{{{{\left( {2x - 1} \right)}^2}}}}}\]