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Section 7.4 Integration of Rational Functions by Partial Fractions
Objectives
Use polynomial long division to write improper rational functions as sums of proper rational functions
Decompose a proper rational function as a sum of partial fractions
Use partial fraction decomposition to compute integrals of rational functions whose denominators are composed of factors of the following forms: distinct linear, repreated, linear, distinct irreducible quadratic, repeated irreducible quadratic
Subsection 7.4.1 Before Class
https://mymedia.ou.edu/media/7.4-1/1_otoe669bFigure 63. Pre-Class Video 1https://mymedia.ou.edu/media/7.4-2/1_60i4tlqfFigure 64. Pre-Class Video 2
Subsubsection 7.4.1.1 Distinct Linear Factors
Example 7.4.1 .
Find \(\ds \int\dfrac{1}{x^2-6x-7}\, dx\)
Why can we not use any of our previous methods on this problem?
Before we can solve this, think about adding the fractions \(\dfrac{1}{2} + \dfrac{1}{3}\text{.}\) What do you have to do in order to add these two fractions?
How could you decompose the answer from above into two fractions? Think about part (b)...
Use your thought process in part (c) to decompose \(f(x)\) into two fractions.
Integrate the resulting decomposition.
Solution .
No appropriate pattern
You need to get a common denominator:
\begin{equation*}
\dfrac{1}{2}\cdot\lrpar{\dfrac{3}{3}} + \dfrac{1}{3}\cdot \lrpar{\dfrac{3}{3}} = \dfrac{3}{6} + \dfrac{2}{6} = \dfrac{5}{6}
\end{equation*}
Write \(\dfrac{5}{6} = \dfrac{A}{2} + \dfrac{B}{3}\)
Write
\(x^2-6x-7 =(x-7)(x+1)\text{.}\) This gives the setup
\begin{equation*}
\dfrac{1}{x^2-6x-7} = \dfrac{1}{(x-7)(x+1)} = \dfrac{A}{x-7} + \dfrac{B}{x+1}
\end{equation*}
Combining fractions we get
\begin{equation*}
\dfrac{1}{(x-7)(x+1)} = \dfrac{A(x+1) + B(x-7)}{(x-7)(x+1)}
\end{equation*}
After simplifying and matching coefficients, we have the system
\begin{equation*}
\begin{cases}
A + B & = 0\\
A-7B & = 1
\end{cases}
\end{equation*}
Solving the system gives
\(A = \dfrac{1}{8}\) and
\(B =-\dfrac{1}{8}\text{,}\) so that our decomposition is
\begin{equation*}
\dfrac{1}{(x-7)(x+1)} = \dfrac{1/8}{x-7} + \dfrac{-1/8}{x+1}
\end{equation*}
\(\displaystyle \ds \int \dfrac{1}{x^2-6x-7}\, dx = \int \dfrac{1/8}{x-7} + \dfrac{-1/8}{x+1}\, dx = \dfrac{1}{8}\ln |x-7| - \dfrac{1}{8}\ln |x+1| + C\)
The previous example gives us a process for integrating particular types of rational functions:
Functions With Two Distinct Linear Factors.
Let \(f(x) = \dfrac{P(x)}{Q(x)}\text{,}\) where \(\text{deg}(P) \lt \text{deg}(Q)\text{.}\) If \(Q(x)\) can be written as \(Q(x) = (x-a)(x-b)\text{,}\) then
\begin{equation*}
\dfrac{P(x)}{Q(x)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}
\end{equation*}
Example 7.4.2 .
Integrate \(\ds \int \dfrac{x^3 + x}{x-1}\, dx\)
Solution . \(\ds \int \dfrac{x^3 + x}{x-1}\, dx = \dfrac{1}{3}x^3+\dfrac{1}{2}x^2 + 2x + 2\ln |x-1| + C\)
Example 7.4.3 .
Integrate \(\ds \int \dfrac{1}{x^2-a^2}\, dx\) using the method of partial fractions.
Solution . \(\ds \int \dfrac{1}{x^2-a^2}\, dx = \dfrac{1}{2a}\ln |x-a| - \dfrac{1}{2a}\ln |x+a| + C\)
Example 7.4.4 .
Integrate \(\ds \int \dfrac{x^2+2x-1}{2x^3+3x^2-2x}\, dx\)
Solution . \(\ds \int \dfrac{x^2+2x-1}{2x^3+3x^2-2x}\, dx = \dfrac{1}{2}\ln |x| + \dfrac{7}{6}\ln |2x-1| - \dfrac{1}{10}\ln |x+2| + C\)
The previous example allows us to generalize our boxed comment from earlier:
Functions With \(n\) Distinct Linear Factors.
Let \(f(x) = \dfrac{P(x)}{Q(x)}\text{,}\) where \(\text{deg}(P) \lt \text{deg}(Q)\text{.}\) If \(Q(x)\) can be written as
\begin{equation*}
Q(x) = (x-x_1)(x-x_2)\cdots(x-x_n)
\end{equation*}
where each \(x_i\) is distinct, then
\begin{equation*}
\dfrac{P(x)}{Q(x)} = \dfrac{A_1}{x-x_1} + \dfrac{A_2}{x-x_2} + \cdots + \dfrac{A_n}{x-x_n}
\end{equation*}
Subsection 7.4.2 Pre-Class Activities
Example 7.4.5 .
Write any questions you have from the videos in this space.
Example 7.4.6 .
Find the general antiderivative of the function \(f(x) = \dfrac{x^4}{x-1}\)
Solution . \(\ds \int f(x)\, dx = \dfrac{1}{4}x^4 + \dfrac{1}{3}x^3 + \dfrac{1}{2}x^2 + x - \dfrac{25}{12} + \ln |x-1| + C\)
Example 7.4.7 .
Integrate \(\ds \int \dfrac{ax}{x^2-bx}\, dx\)
Solution . \(\ds \int \dfrac{ax}{x^2-bx}\, dx = a\ln |x-b| + C\)
Example 7.4.8 .
Integrate \(\ds \int \dfrac{x}{(x-1)(x+1)(2x+1)}\, dx\)
Solution . \(\ds \int \dfrac{x}{(x-1)(x+1)(2x+1)}\, dx = \dfrac{1}{6}\ln |x-1| - \dfrac{1}{2}\ln |x+1| + \dfrac{1}{3}\ln |2x + 1| + C\)
Subsection 7.4.3 In Class
Subsubsection 7.4.3.1 Repeated Linear Factors
Example 7.4.9 .
Find \(\ds \int \dfrac{x^6}{x^2-4}\, dx\)
Solution . \(\ds \int \dfrac{x^6}{x^2-4}\, dx = \dfrac{1}{5}x^5 + \dfrac{4}{3}x^3 + 16x -16\ln |x+2| + 16 \ln |x-2| + C\)
Functions with Repeated Linear Factors.
Let \(f(x) = \dfrac{P(x)}{Q(x)}\text{,}\) where \(\text{deg}(P) \lt \text{deg}(Q)\text{.}\) If \(Q\) can be written as
\begin{equation*}
Q(x) =(x-x_1)^{m_1}(x-x_2)^{m_2}\cdots (x-x_n)^{m_n}
\end{equation*}
where each \(x_i\) is distinct, then
\begin{equation*}
\dfrac{P(x)}{Q(x)} = \dfrac{A_1}{x-x_1} + \dfrac{A_2}{(x-x_1)^2} +\cdots + \dfrac{A_{m_1}}{(x-x_1)^{m_1}} + \dfrac{B_1}{x-x_2} + \dfrac{B_2}{(x-x_2)^2} + \cdots
\end{equation*}
Example 7.4.10 .
Evaluate \(\ds \int \dfrac{x}{(x-1)^2}\, dx\)
Solution . \(\ds \int \dfrac{x}{(x-1)^2}\, dx = \ln |x-1| - \dfrac{1}{x-1} + C\)
Example 7.4.11 .
Find \(\ds \int \dfrac{x^4-2x^2+4x+1}{x^3-x^2-x+1}\, dx\)
Solution . \(\ds \int \dfrac{x^4-2x^2+4x+1}{x^3-x^2-x+1}\, dx = \dfrac{1}{2}x^2 + x - \ln |x+1| + \ln |x-1| - \dfrac{2}{x-1} + C\)
Example 7.4.12 .
Evaluate \(\ds \int \dfrac{x^3+4x^2+x-1}{x^3+x^2}\, dx\)
Solution . \(\ds \int \dfrac{x^3+4x^2+x-1}{x^3+x^2}\, dx = x+2\ln |x| + \dfrac{1}{x} + \ln |x+1| + C\)
Subsubsection 7.4.3.2 Irreducible Quadratic Factors
Definition 7.4.13 . Irreducible Quadratic.
The polynomial \(ax^2 + bx + c\) is said to be irreducible if
\begin{equation*}
b^2-4ac \lt 0
\end{equation*}
Functions with An Irreducible Quadratic Factor.
Let \(f(x) = \dfrac{P(x)}{Q(x)}\text{,}\) where \(\text{deg}(P) \lt \text{deg}(Q)\text{.}\) If \(Q(x)\) has an irreducible quadratic factor, then the decomposition of \(f(x) = \dfrac{P(x)}{Q(x)}\) will have a term of the form
\begin{equation*}
\dfrac{Ax + B}{ax^2 + bx + c}
\end{equation*}
Example 7.4.14 .
Evaluate \(\ds \int \dfrac{2x^2-x+4}{x^3 + 4x}\, dx\)
Solution . \(\ds \int \dfrac{2x^2-x+4}{x^3 + 4x}\, dx = \ln |x| + \dfrac{1}{2}\ln (x^2 + 4) - \dfrac{1}{2}\inv{\tan}\lrpar{\dfrac{x}{2}} + C\)
Example 7.4.15 .
Evaluate \(\ds \int \dfrac{x-1}{4x^2-4x+3}\, dx\)
Solution . \(\ds \int \dfrac{x-1}{4x^2-4x+3}\, dx = \dfrac{1}{8}\ln (4x^2-4x + 3) -\dfrac{\sqrt{2}}{8}\inv{\tan}\lrpar{\dfrac{2x-1}{\sqrt{2}}} + C\)
Subsubsection 7.4.3.3 Functions with Repeated Irreducible Quadratic Factors
Let \(f(x) = \dfrac{P(x)}{Q(x)}\text{,}\) where \(\text{deg}(P) \lt \text{deg}(Q)\text{.}\) If \(Q(x)\) has an irreducible quadratic factor of the form \((ax^2 + bx + c)^r\text{,}\) then the decomposition of \(f(x) = \dfrac{P(x)}{Q(x)}\) will have a term of the form
\begin{equation*}
\dfrac{A_1x + B_1}{ax^2 + bx + c} + \dfrac{A_2x+B_2}{(ax^2+bx+c)^2} + \cdots + \dfrac{A_rx + B_r}{(ax^2+bx+c)^r}
\end{equation*}
Example 7.4.16 .
Write the partial fraction decomposition for the function \(f(x) = \dfrac{x}{x(x+1)(x^2+x+1)(x^2+3)^3}\)
Solution . \(\dfrac{x}{x(x+1)(x^2+x+1)(x^2+3)^3} = \dfrac{A}{x} + \dfrac{B}{x+1} + \dfrac{Cx+D}{x^2+x+1} + \dfrac{Ex+F}{x^2 + 3} + \dfrac{Gx+H}{(x^2+3)^2} + \dfrac{Ix + J}{(x^2+3)^3}\)
Example 7.4.17 .
Compute \(\ds \int \dfrac{x^2+x+1}{(x^2+1)^2}\, dx\)
Solution . \(\ds \int \dfrac{x^2+x+1}{(x^2+1)^2}\, dx = \inv{\tan}(x) - \dfrac{1}{2}\cdot \dfrac{1}{x^2+1} + C\)
Example 7.4.18 .
Compute \(\ds \int \dfrac{x^3+6x-2}{x^4+6x^2}\, dx\)
Solution . \(\ds \int \dfrac{x^3+6x-2}{x^4+6x^2}\, dx = \ln |x| + \dfrac{1}{3}\cdot \dfrac{1}{x} + \dfrac{1}{3\sqrt{6}}\inv{\tan}\lrpar{\dfrac{x}{\sqrt{6}}} + C\)
Example 7.4.19 .
Compute \(\ds\int \dfrac{4x}{x^3 + x^2 + x + 1}\, dx\)
Solution . \(\ds\int \dfrac{4x}{x^3 + x^2 + x + 1}\, dx = \ln (x^2 + 1) + 2\inv{\tan}(x) - 2\ln |x+1| + C\)
Example 7.4.20 .
Use substitution to evaluate \(\ds \int \dfrac{dx}{x\sqrt{x-2}}\)
Solution . \(\ds \int \dfrac{dx}{x\sqrt{x-2}} = \sqrt{2}\inv{\tan}\lrpar{\dfrac{\sqrt{x-2}}{\sqrt{2}}}\)
Example 7.4.21 .
Use substitution to evaluate \(\ds \int \dfrac{1}{(1+\sqrt{x})^2}\, dx\)
Solution . \(\ds \int \dfrac{1}{(1+\sqrt{x})^2}\, dx = \dfrac{2}{\sqrt{x} + 1} + 2\ln |\sqrt{x} + 1| + C\)
Example 7.4.22 .
Compute \(\ds \int \dfrac{e^x}{(e^x-2)(e^{2x}+1)}\, dx\)
Solution . \(\ds \int \dfrac{e^x}{(e^x-2)(e^{2x}+1)}\, dx = \dfrac{1}{5}\ln (e^x-2) -\dfrac{1}{10}\ln (e^{2x} + 1) - \dfrac{2}{5}\inv{\tan} (e^x) + C\)
Subsection 7.4.4 After Class Activities
Example 7.4.23 .
Evaluate the following integrals:
\(\displaystyle \ds \int_0^1 \dfrac{x-4}{x^2-5x+6}\, dx\)
\(\displaystyle \ds \int \dfrac{1}{(t^2-1)^2}\, dt\)
\(\displaystyle \ds \int \dfrac{10}{(x-1)(x^2+9)}\, dx\)
\(\displaystyle \ds \int \dfrac{x^4 + 9x^2 + x + 2}{x^2 + 9}\, dx\)
Solution .
\(\displaystyle \ds \int_0^1 \dfrac{x-4}{x^2-5x+6}\, dx = \ln 3 - \ln 8\)
\(\displaystyle \ds \int \dfrac{1}{(t^2-1)^2}\, dt = -\dfrac{1}{2}\cdot \dfrac{x}{x^2-1} - \dfrac{1}{4}\ln|x-1| + \dfrac{1}{4}\ln |x+1| + C\)
\(\displaystyle \ds \int \dfrac{10}{(x-1)(x^2+9)}\, dx = \dfrac{1}{2}\ln (x^2 + 9) + \ln |x-1| - \dfrac{1}{3}\inv{\tan}\lrpar{\dfrac{x}{3}} + C\)
\(\displaystyle \ds \int \dfrac{x^4 + 9x^2 + x + 2}{x^2 + 9}\, dx = \dfrac{1}{3}x^3 + \dfrac{1}{2}\ln (x^2 + 9) + \dfrac{2}{3}\inv{\tan}\lrpar{\dfrac{x}{3}} + C\)
