Section 7.4 : Partial Fractions
9. Evaluate the integral \( \displaystyle \int{{\frac{{6{x^2} - 3x}}{{\left( {x - 2} \right)\left( {x + 4} \right)}}\,dx}}\).
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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 2Now 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 3The “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 4We can now do the integral.
\[\int{{\frac{{6{x^2} - 3x}}{{\left( {x - 2} \right)\left( {x + 4} \right)}}\,dx}} = \int{{6 + \frac{3}{{x - 2}} - \frac{{18}}{{x + 4}}\,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{6x + 3\ln \left| {x - 2} \right| - 18\ln \left| {x + 4} \right| + c}}\]