Calculus II - Partial Fractions

Hide Mobile Notice

Mobile Notice

You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 7.4 : Partial Fractions

Back to Problem List

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

Show All Steps Hide All Steps

Start Solution

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.

Show Step 3

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.