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### Section 7.4 : Partial Fractions

8. Evaluate the integral $$\displaystyle \int{{\frac{{8 + t + 6{t^2} - 12{t^3}}}{{\left( {3{t^2} + 4} \right)\left( {{t^2} + 7} \right)}}\,dt}}$$.

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In this case the denominator is already factored and so we can go straight to the form of the partial fraction decomposition for the integrand.

$\frac{{8 + t + 6{t^2} - 12{t^3}}}{{\left( {3{t^2} + 4} \right)\left( {{t^2} + 7} \right)}} = \frac{{At + B}}{{3{t^2} + 4}} + \frac{{Ct + D}}{{{t^2} + 7}}$ Show Step 2

Setting the numerators equal gives,

\begin{align*}8 + t + 6{t^2} - 12{t^3} & = \left( {At + B} \right)\left( {{t^2} + 7} \right) + \left( {Ct + D} \right)\left( {3{t^2} + 4} \right)\\ & = \left( {A + 3C} \right){t^3} + \left( {B + 3D} \right){t^2} + \left( {7A + 4C} \right)t + 7B + 4D\end{align*}

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.

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Now, setting the coefficients equal gives the following system.

\begin{align*}{{t^3}:} & \hspace{0.25in} & A + 3C = & - 12\\{{t^2}:} & \hspace{0.25in} & B + 3D = & \, 6\\{{t^1}:} & \hspace{0.25in} & 7A + 4C = &\, 1\\{{t^0}:} & \hspace{0.25in} & 7B + 4D = &\, 8\end{align*}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{aligned} & {A = 3}\\ & {B = 0}\\ & {C = - 5}\\ & {D = 2}\end{aligned}

The partial fraction form of the integrand is then,

$\frac{{8 + t + 6{t^2} - 12{t^3}}}{{\left( {3{t^2} + 4} \right)\left( {{t^2} + 7} \right)}} = \frac{{3t}}{{3{t^2} + 4}} + \frac{{ - 5t + 2}}{{{t^2} + 7}}$ Show Step 4

We can now do the integral.

\begin{align*}\int{{\frac{{8 + t + 6{t^2} - 12{t^3}}}{{\left( {3{t^2} + 4} \right)\left( {{t^2} + 7} \right)}}\,dt}} & = \int{{\frac{{3t}}{{3{t^2} + 4}} + \frac{{ - 5t + 2}}{{{t^2} + 7}}\,dt}}\\ & = \int{{\frac{{3t}}{{3{t^2} + 4}} - \frac{{5t}}{{{t^2} + 7}} + \frac{2}{{{t^2} + 7}}\,dt}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{2}\ln \left| {3{t^2} + 4} \right| - \frac{5}{2}\ln \left| {{t^2} + 7} \right| + \frac{2}{{\sqrt 7 }}{{\tan }^{ - 1}}\left( {\frac{t}{{\sqrt 7 }}} \right) + c}}\end{align*}

Note that the first and second integrations needed the substitutions $$u = 3{t^2} + 4$$ and $$u = {t^2} + 7$$ respectively while the third needed the formula provided in the notes.