Calculus II - Partial Fractions

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Section 7.4 : Partial Fractions

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6. Evaluate the integral $\displaystyle \int{{\frac{{4x - 11}}{{{x^3} - 9{x^2}}}\,dx}}$ .

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To 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 2

Setting the numerators equal gives,

$4x - 11 = Ax\left( {x - 9} \right) + B\left( {x - 9} \right) + C\,{x^2}$ Show Step 3

We 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 4

We can now do the integral.

$\int{{\frac{{4x - 11}}{{{x^2}\left( {x - 9} \right)}}\,dx}} = \int{{\frac{{ - \frac{{25}}{{81}}}}{x} + \frac{{\frac{{11}}{9}}}{{{x^2}}} + \frac{{\frac{{25}}{{81}}}}{{x - 9}}\,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{25}}{{81}}\ln \left| x \right| - \frac{{\frac{{11}}{9}}}{x} + \frac{{25}}{{81}}\ln \left| {x - 9} \right| + c}}$