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
6. Evaluate the integral \( \displaystyle \int{{\frac{{4x - 11}}{{{x^3} - 9{x^2}}}\,dx}}\).
Show All Steps Hide All Steps
Start SolutionTo get the problem started off we need the form of the partial fraction decomposition of the integrand. However, in order to get this, we’ll need to factor the denominator.
\[\int{{\frac{{4x - 11}}{{{x^3} - 9{x^2}}}\,dx}} = \int{{\frac{{4x - 11}}{{{x^2}\left( {x - 9} \right)}}\,dx}}\]The form of the partial fraction decomposition for the integrand is then,
\[\frac{{4x - 11}}{{{x^2}\left( {x - 9} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x - 9}}\] Show Step 2Setting the numerators equal gives,
\[4x - 11 = Ax\left( {x - 9} \right) + B\left( {x - 9} \right) + C\,{x^2}\] Show Step 3We can use the “trick” discussed in the notes to easily get two of the coefficients and then we can just pick another value of \(x\) to get the third so let’s do that. Here is that work.
\[\begin{align*}x = & \, 0 : & - 11 = & - 9B\\ & & & \\x = & \,9 : & 25 = &\, 81C\\ & & & \\x = & \,1 : & - 7 = & - 8A - 8B + C = - 8A - \frac{{767}}{{81}}\end{align*}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{aligned} & {A = - \frac{{25}}{{81}}}\\ & {B = \frac{{11}}{9}}\\ & {C = \frac{{25}}{{81}}}\end{aligned}\]The partial fraction form of the integrand is then,
\[\frac{{4x - 11}}{{{x^2}\left( {x - 9} \right)}} = \frac{{ - \frac{{25}}{{81}}}}{x} + \frac{{\frac{{11}}{9}}}{{{x^2}}} + \frac{{\frac{{25}}{{81}}}}{{x - 9}}\] Show Step 4We can now do the integral.
\[\int{{\frac{{4x - 11}}{{{x^2}\left( {x - 9} \right)}}\,dx}} = \int{{\frac{{ - \frac{{25}}{{81}}}}{x} + \frac{{\frac{{11}}{9}}}{{{x^2}}} + \frac{{\frac{{25}}{{81}}}}{{x - 9}}\,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{25}}{{81}}\ln \left| x \right| - \frac{{\frac{{11}}{9}}}{x} + \frac{{25}}{{81}}\ln \left| {x - 9} \right| + c}}\]