Section 7.4 : Partial Fractions
10. Evaluate the integral \( \displaystyle \int{{\frac{{2 + {w^4}}}{{{w^3} + 9w}}\,dw}}\).
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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 the degree of the numerator is 4 and the degree of the denominator is 3.
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,
\[\frac{{2 + {w^4}}}{{{w^3} + 9w}} = w + \frac{{2 - 9{w^2}}}{{w\left( {{w^2} + 9} \right)}}\] Show Step 2Now we can do the partial fractions on the second term. Here is the form of the partial fraction decomposition.
\[\frac{{2 - 9{w^2}}}{{w\left( {{w^2} + 9} \right)}} = \frac{A}{w} + \frac{{Bw + C}}{{{w^2} + 9}}\]Setting the numerators equal gives,
\[2 - 9{w^2} = A\left( {{w^2} + 9} \right) + w\left( {Bw + C} \right) = \left( {A + B} \right){w^2} + Cw + 9A\]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 3Now, setting the coefficients equal gives the following system.
\[\begin{align*}{{w^2}:} & \hspace{0.25in} & A + B = & - 9\\ & & & \\{{w^1}:} & \hspace{0.25in} & C = & \,0\\ & & & \\{{w^0}:} & \hspace{0.25in} & 9A = &\, 2\end{align*}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{aligned} & {A = \frac{2}{9}}\\ & {B = - \frac{{83}}{9}}\\ & {C = 0}\end{aligned}\]The partial fraction form of the second term is then,
\[\frac{{2 - 9{w^2}}}{{w\left( {{w^2} + 9} \right)}} = \frac{{\frac{2}{9}}}{w} - \frac{{\frac{{83}}{9}w}}{{{w^2} + 9}}\] Show Step 4We can now do the integral.
\[\int{{\frac{{2 + {w^4}}}{{{w^3} + 9w}}\,dw}} = \int{{w + \frac{{\frac{2}{9}}}{w} - \frac{{\frac{{83}}{9}w}}{{{w^2} + 9}}\,dw}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{2}{w^2} + \frac{2}{9}\ln \left| w \right| - \frac{{83}}{{18}}\ln \left| {{w^2} + 9} \right| + c}}\]