9. Partial Fractions

c2. Integral Examples

On previous pages, we found the partial fraction expansions for several rational functions. Here are their integrals:
The first four integrals just need \(u\)-substitutions.

Non-Repeated Linear Factors

Compute:   \(\displaystyle \int \dfrac{x^2-11x+12}{x^3-5x^2+6x}\,dx\)   given that   \(\dfrac{x^2-11x+12}{x^3-5x^2+6x} =\dfrac{2}{x}+\dfrac{3}{x-2}-\dfrac{4}{x-3}\).
How are the coefficients found?

\[\begin{aligned} \int \dfrac{x^2-11x+12}{x^3-5x^2+6x}\,dx &=\int \dfrac{2}{x}\,dx+\int \dfrac{3}{x-2}\,dx-\int \dfrac{4}{x-3}\,dx \\ &=2\ln|x|+3\ln|x-2|-4\ln|x-3|+C \end{aligned}\]

Repeated Linear Factors

Compute:   \(\displaystyle \int \dfrac{x^2+1}{x^4+3x^3+3x^2+x}\,dx\)   given that   \(\displaystyle \dfrac{x^2+1}{x^4+3x^3+3x^2+x} =\dfrac{1}{x}-\dfrac{1}{x+1}-\dfrac{2}{(x+1)^3}\)
How are the coefficients found?

\[\begin{aligned} \int \dfrac{x^2+1}{x^4+3x^3+3x^2+x}\,dx &=\int \dfrac{1}{x}\,dx-\int \dfrac{1}{x+1}\,dx -\int \dfrac{2}{(x+1)^3}\,dx \\ &=\ln|x|-\ln|x+1|+\dfrac{1}{(x+1)^2}+C \end{aligned}\]

Non-Repeated Quadratic Factors

Compute:   \(\displaystyle \int \dfrac{13}{x^3-4x^2+13x}\,dx\)   given that   \(\dfrac{13}{x^3-4x^2+13x}=\dfrac{1}{x}+\dfrac{-(x-2)+2}{(x-2)^2+3^2}\)
How are the coefficients found?

The quadratic term must be broken up: \[ \int \dfrac{13}{x^3-4x^2+13x}\,dx =\int \dfrac{1}{x}\,dx-\int \dfrac{(x-2) }{(x-2)^2+3^2}\,dx+2\int \dfrac{1}{(x-2)^2+3^2}\,dx \] The first integral is \[ \int \dfrac{1}{x}\,dx=\ln|x| \] In the second integral, we make the substitution \(u=(x-2)^2+3^2\). Then \(du=2(x-2)\,dx\) and \(\dfrac{1}{2}\,du=(x-2)\,dx\) and so \[\begin{aligned} \int \dfrac{(x-2) }{(x-2)^2+3^2}\,dx &=\dfrac{ 1}{2}\int \dfrac{1}{u}\,du=\dfrac{1}{2}\ln|u|=\dfrac{1}{2} \ln|(x-2)^2+3^2| \\ &=\dfrac{1}{2}\ln|x^2-4x+13| \end{aligned}\] In the third integral, we make the substitution \(x-2=3\tan\theta\). Then \(dx=3\sec^2\theta\,d\theta\). So \[\begin{aligned} \int \dfrac{1}{(x-2)^2+3^2}\,dx &=\int \dfrac{1}{3^2\tan^2\theta+3^2}\,3\sec^2\theta\,d\theta =\dfrac{1}{3}\int \dfrac{\sec^2\theta}{\tan^2\theta+1}\,d\theta \\ &=\dfrac{1}{3}\int 1\,d\theta=\dfrac{1}{3}\theta=\dfrac{1}{3}\arctan \left(\dfrac{x-2}{3}\right) \end{aligned}\] Putting these together, we get \[ \int \dfrac{13}{x^3-4x^2+13x}\,dx=\ln|x| -\dfrac{1}{2}\ln|x^2-4x+13|+\dfrac{2}{3}\arctan\left(\dfrac{x-2}{3}\right)+C \]

Repeated Quadratic Factors

This example is probably harder than anything you will be asked to do in this course. It is included for completeness.

Compute:   \(\displaystyle \int \dfrac{100}{(x-1) \left((x-2)^2+3^2\right)^2}\,dx\)   given that   \(\dfrac{100}{(x-1) \left((x-2)^2+3^2\right)^2} =\dfrac{1}{x-1}+\dfrac{-(x-2)+1}{(x-2)^2+3^2}+\dfrac{-10(x-2)+10}{\left((x-2)^2+3^2\right)^2}\)
How are the coefficients found?

Both quadratic terms must be broken up: \[\begin{aligned} \int &\dfrac{100}{(x-1) \left((x-2)^2+3^2\right)^2}\,dx \\ &=\int \dfrac{1}{x-1}\,dx-\int \dfrac{(x-2) }{(x-2)^2+3^2}\,dx +\int \dfrac{1}{(x-2)^2+3^2}\,dx \\ & \quad -10\int \dfrac{(x-2) }{\left((x-2)^2+3^2\right)^2}\,dx +10\int \dfrac{1}{\left((x-2)^2+3^2\right)^2}\,dx \end{aligned}\] The first integral is \(\ln|x-1|\). The second and fourth integrals may be done with the substitution \(u=(x-2)^2+3^2\). Then \(du=2(x-2)\,dx\) and so \((x-2)\,dx=\dfrac{1}{2}\,du\). Thus \[ \int \dfrac{(x-2) }{(x-2)^2+3^2}\,dx =\dfrac{1}{2}\int \dfrac{1}{u}\,du =\dfrac{1}{2}\ln|u| =\dfrac{1}{2}\ln\left|(x-2)^2+3^2\right| \] \[ \int \dfrac{(x-2) }{\left((x-2)^2+3^2\right)^2}\,dx =\dfrac{1}{2}\int \dfrac{1}{u^2}\,du =-\dfrac{1}{2u} =\dfrac{-1}{2\left((x-2)^2+3^2\right) } \] The third and fifth integrals may be done with the substitution \(x-2=3\tan\theta\). Then \(dx=3\sec^2\theta\,d\theta\). Thus the third integral is: \[\begin{aligned} \int &\dfrac{1}{(x-2)^2+3^2}\,dx =\int \dfrac{1}{3^2\tan^2\theta+3^2}\,3\sec^2\theta\,d\theta \\ &=\dfrac{1}{3}\int \dfrac{\sec^2\theta}{\tan^2\theta+1}\,d\theta =\dfrac{1}{3}\int 1\,d\theta =\dfrac{1}{3}\theta \\ &=\dfrac{1}{3}\arctan \left(\dfrac{x-2}{3}\right) \end{aligned}\] And the fifth integral is: \[\begin{aligned} \int &\dfrac{1}{\left((x-2)^2+3^2\right)^2}\,dx =\int \dfrac{1}{\left(3^2\tan^2\theta+3^2\right)^2}\,3\sec^2\theta\,d\theta \\ &=\dfrac{1}{27}\int \dfrac{\sec^2\theta}{\left(\sec^2\theta\right)^2}\,d\theta =\dfrac{1}{27}\int \cos^2\theta\,d\theta =\dfrac{1}{27}\int \dfrac{1+\cos(2\theta) }{2}\,d\theta \\ &=\dfrac{1}{54}\left(\theta+\dfrac{\sin(2\theta) }{2}\right) =\dfrac{1}{54}(\theta+\sin\theta\cos\theta) \end{aligned}\] Now since \(\tan\theta=\dfrac{x-2}{3}\), we have \(\sin\theta=\dfrac{x-2}{\sqrt{(x-2)^2+3^2}}\) and \(\cos\theta=\dfrac{3}{\sqrt{(x-2)^2+3^2}}\). So the fifth integral becomes \[ \int \dfrac{1}{\left((x-2)^2+3^2\right)^2}\,dx =\dfrac{1}{54}\left(\arctan \left(\dfrac{x-2}{3}\right) +\dfrac{3(x-2) }{(x-2)^2+3^2}\right) \] Putting these together, we get \[\begin{aligned} \int &\dfrac{100}{(x-1) \left((x-2)^2+3^2\right)^2}\,dx \\ &=\int \dfrac{1}{x-1}\,dx-\int \dfrac{(x-2) }{(x-2)^2+3^2}\,dx +\int \dfrac{1}{(x-2)^2+3^2}\,dx \\ & \quad -10\int \dfrac{(x-2) }{\left((x-2)^2+3^2\right)^2}\,dx +10\int \dfrac{1}{\left((x-2)^2+3^2\right)^2}\,dx \\ &=\ln|x-1|-\dfrac{1}{2}\ln|(x-2)^2+3^2| +\dfrac{1}{3}\arctan \left(\dfrac{x-2}{3}\right) \\ & \quad -10\left[ \dfrac{-1}{2\left((x-2)^2+3^2\right) }\right] +10\left[ \dfrac{1}{54}\left(\arctan \left(\dfrac{x-2}{3}\right) +\dfrac{3(x-2) }{(x-2)^2+3^2}\right) \right]+C \\ &=\ln|x-1|-\dfrac{1}{2}\ln\left|(x-2)^2+3^2\right| +\dfrac{14}{27}\arctan \left(\dfrac{x-2}{3}\right) \\ & \quad +\dfrac{5}{(x-2)^2+3^2}+\dfrac{5}{9}\dfrac{x-2}{(x-2)^2+3^2}+C \end{aligned}\] The Bad News: This is a lot of work and you need to do it slowly and carefully.
The Good News: This is as hard as it gets.