Paul's Online Notes
Home / Calculus II / Integration Techniques / Partial Fractions
Show Mobile Notice Show All Notes Hide All Notes
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 views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 7.4 : Partial Fractions

9. Evaluate the integral $$\displaystyle \int{{\frac{{6{x^2} - 3x}}{{\left( {x - 2} \right)\left( {x + 4} \right)}}\,dx}}$$.

Show All Steps Hide All Steps

Hint : Pay attention to the degree of the numerator and denominator!
Start Solution

Remember that we can only do partial fractions on a rational expression if the degree of the numerator is less than the degree of the denominator. In this case both the numerator and denominator are both degree 2. This can be easily seen if we multiply the denominator out.

$\frac{{6{x^2} - 3x}}{{\left( {x - 2} \right)\left( {x + 4} \right)}} = \frac{{6{x^2} - 3x}}{{{x^2} + 2x - 8}}$

So, the first step is to do long division (we’ll leave it up to you to check our Algebra skills for the long division) to get,

$\frac{{6{x^2} - 3x}}{{\left( {x - 2} \right)\left( {x + 4} \right)}} = 6 + \frac{{48 - 15x}}{{\left( {x - 2} \right)\left( {x + 4} \right)}}$ Show Step 2

Now we can do the partial fractions on the second term. Here is the form of the partial fraction decomposition.

$\frac{{48 - 15x}}{{\left( {x - 2} \right)\left( {x + 4} \right)}} = \frac{A}{{x - 2}} + \frac{B}{{x + 4}}$

Setting the numerators equal gives,

$48 - 15x = A\left( {x + 4} \right) + B\left( {x - 2} \right)$ Show Step 3

The “trick” will work here easily enough so here is that work.

\begin{align*}{x = - 4:} & \hspace{0.25in} & 108 = & - 6B\\{x = 2:} & \hspace{0.25in} & 18 = & \, 6A\end{align*}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{aligned} & {A = 3}\\ & {B = - 18}\end{aligned}

The partial fraction form of the second term is then,

$\frac{{48 - 15x}}{{\left( {x - 2} \right)\left( {x + 4} \right)}} = \frac{3}{{x - 2}} - \frac{{18}}{{x + 4}}$ Show Step 4

We can now do the integral.

$\int{{\frac{{6{x^2} - 3x}}{{\left( {x - 2} \right)\left( {x + 4} \right)}}\,dx}} = \int{{6 + \frac{3}{{x - 2}} - \frac{{18}}{{x + 4}}\,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{6x + 3\ln \left| {x - 2} \right| - 18\ln \left| {x + 4} \right| + c}}$