Calculus II - Partial Fractions

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Section 7.4 : Partial Fractions

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2. Evaluate the integral \( \displaystyle \int{{\frac{{8 - 3t}}{{10{t^2} + 13t - 3}}\,dt}}\).

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To get the problem started off we need the form of the partial fraction decomposition of the integrand. However, in order to get this, we’ll need to factor the denominator.

\[\int{{\frac{{8 - 3t}}{{10{t^2} + 13t - 3}}\,dt}} = \int{{\frac{{8 - 3t}}{{\left( {2t + 3} \right)\left( {5t - 1} \right)}}\,dt}}\]

The form of the partial fraction decomposition for the integrand is then,

\[\frac{{8 - 3t}}{{10{t^2} + 13t - 3}} = \frac{A}{{2t + 3}} + \frac{B}{{5t - 1}}\] Show Step 2

Setting the numerators equal gives,

\[8 - 3t = A\left( {5t - 1} \right) + B\left( {2t + 3} \right)\] Show Step 3

We can use the “trick” discussed in the notes to easily get the coefficients in this case so let’s do that. Here is that work.

\[\begin{align*}t = & \frac{1}{5} \,\,\,\,\,\,\, : & \frac{{37}}{5} & = \frac{{17}}{5}B\\ & & & \\t = & - \frac{3}{2}: & \frac{{25}}{2} & = - \frac{{17}}{2}A\end{align*}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{aligned}&{A = - \frac{{25}}{{17}}}\\ & \\&{B = \frac{{37}}{{17}}}\end{aligned}\]

The partial fraction form of the integrand is then,

\[\frac{{8 - 3t}}{{10{t^2} + 13t - 3}} = \frac{{ - \frac{{25}}{{17}}}}{{2t + 3}} + \frac{{\frac{{37}}{{17}}}}{{5t - 1}}\] Show Step 4

We can now do the integral.

\[\int{{\frac{{8 - 3t}}{{10{t^2} + 13t - 3}}}}\,dt = \int{{\frac{{ - \frac{{25}}{{17}}}}{{2t + 3}} + \frac{{\frac{{37}}{{17}}}}{{5t - 1}}\,dt}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{37}}{{85}}\ln \left| {5t - 1} \right| - \frac{{25}}{{34}}\ln \left| {2t + 3} \right| + c}}\]

Hopefully you are getting good enough with integration that you can do some of these integrals in your head. Be careful however with both of these integrals. When doing these kinds of integrals in our head it is easy to forget about the substitutions that are technically required to do them and then miss the coefficients from the substitutions that need to show up in the answer.