AC Integrating Rational Functions

Section 9.2 Integrating Rational Functions

Subsection 9.2.1 Preview Activity

Question 9.2.1.

For each of the indefinite integrals below, decide whether the integral can be evaluated: immediately because it's a basic integral, using \(u\)-substitution, integration by parts, with multiple techniques, or if none of those will work.

\(\displaystyle \int \frac{1}{1+x^2} \, dx\)

Basic Integral
u-Substitution
Integration by Parts
A combination of techniques
None of the above

\(\displaystyle \int \frac{x}{1+x^2} \, dx\)

Basic Integral
u-Substitution
Integration by Parts
A combination of techniques
None of the above

\(\displaystyle \int \frac{2x+3}{1+x^2} \, dx\)

Basic Integral
u-Substitution
Integration by Parts
A combination of techniques
None of the above

\(\displaystyle \int \frac{e^x}{1+(e^x)^2} \, dx\)

Basic Integral
u-Substitution
Integration by Parts
A combination of techniques
None of the above

\(\displaystyle \int \frac{1}{x^2-1} \, dx\)

Basic Integral
u-Substitution
Integration by Parts
A combination of techniques
None of the above

Hint.

Note that \(\displaystyle \frac{2x+3}{1+x^2}\) can be rewritten as \(\displaystyle \frac{2x}{1+x^2} +\frac{3}{1+x^2}\text{.}\)

Question 9.2.2.

First, note that we can use algebra to combine \(\frac{-5}{x+1}+\frac{6}{x-1}\) into one fraction by finding a common denominator.

\begin{equation*} \frac{-5}{x+1}+\frac{6}{x-1} = \frac{-5}{x+1}\cdot\frac{x-1}{x-1}+\frac{6}{x-1}\cdot \frac{x+1}{x+1}=\frac{x+11}{x^{2}-1} \end{equation*}

Also note that we know how to integrate

\(\displaystyle \int \frac{-5}{x+1} \, dx =\) and

\(\displaystyle \int \frac{6}{x-1} \, dx =\)

Which means we can integrate both sides of \(\frac{-5}{x+1}+\frac{6}{x-1} = \frac{x+11}{x^{2}-1}\) to get

\(\displaystyle \int \frac{x+11}{x^{2}-1} \, dx =\)

Hint.

Remember that you really should use absolute value instead of parentheses when you integrate to get \(\ln\text{.}\) Sometimes WeBWorK doesn't care about the difference, but sometimes it does. For example, \(\int \frac{1}{x+11} \, dx = \ln |x+11| +C\text{,}\) not \(\ln(x+11) +C\text{.}\)

Question 9.2.3.

Find the quotient and remainder using long division for

\begin{equation*} \frac{x^5- x^4+ 3 x^3 - 3 x^2+ 5 x - 8 }{x-1}. \end{equation*}

The quotient is

The remainder is

Question 9.2.4.

Perform the indicated division and write the quotient and remainder in the provided blanks.

\begin{equation*} \frac{3 x^2 - x - 6}{x-1} \end{equation*}

Answer: \(+\) \(/ (x - 1)\)

Exercises 9.2.2 Exercises

1.

Which of the following is the correct form of the partial fraction decomposition of \(\displaystyle \frac{x-1}{x^3+x^2}?\)

\(\displaystyle \displaystyle \frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}\)
\(\displaystyle \displaystyle \frac{Ax+B}{x}+\frac{Cx+D}{x^2}+\frac{Ex+F}{x+1}\)
\(\displaystyle \displaystyle \frac{Ax+B}{x}+\frac{Cx+D}{x^2}+\frac{Ex+F}{x+1}\)
\(\displaystyle \displaystyle \frac{Ax+B}{x}+\frac{Cx+D}{x^2}+\frac{E}{x+1}\)

2.

Which of the following is the correct form of the partial fraction decomposition of \(\displaystyle \frac{x-1}{x^3+x}?\)

\(\displaystyle \displaystyle \frac{A}{x}+\frac{Bx+C}{x^2+1}\)
\(\displaystyle \displaystyle \frac{A}{x}+\frac{B}{x^2+1}\)
\(\displaystyle \displaystyle \frac{Ax+B}{x}+\frac{Cx+D}{x^2+1}\)
\(\displaystyle \displaystyle \frac{Ax+B}{x}+\frac{C}{x^2+1}\)

3.

What is the correct form of the partial fraction decomposition for the following integral?

\begin{equation*} \int \frac{x - 11}{x^3 + 11 x^2 - 12 x} \, dx \end{equation*}

\(\displaystyle \displaystyle \int \left( \frac{A x + B}{x} + \frac{C x + D}{x - 1} + \frac{E x + F}{x + 12} \right)\, dx\)
\(\displaystyle \displaystyle \int \left( \frac{A}{x} + \frac{B x + C}{x - 11} + \frac{D x + E}{x + 1} \right)\, dx\)
\(\displaystyle \displaystyle \int \left( \frac{A}{x} + \frac{B}{x - 11} + \frac{C}{x + 1} \right)\, dx\)
There is no partial fraction decomposition because the denominator does not factor.
There is no partial fraction decomposition yet because there is cancellation.
\(\displaystyle \displaystyle \int \left( \frac{A}{x - 11} + \frac{B}{x - 12} + \frac{C}{x + 1} \right)\, dx\)
There is no partial fraction decomposition yet because long division must be done first.
\(\displaystyle \displaystyle \int \left( \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 12} \right)\, dx\)

4.

What is the correct form of the partial fraction decomposition for the following integral?

\begin{equation*} \int \frac{13 (x^{6} + 8)}{(x - 9)(x^2 - 1)^2 (x^2 + 9)^2} \, dx \end{equation*}

\(\displaystyle \displaystyle \int \left( \frac{A}{x - 9} + \frac{Bx+C}{x^2-1} + \frac{Cx+D}{(x^2-1)^2} + \frac{Ex+F}{x^2 + 9} + \frac{Gx+H}{(x^2 + 9)^2} \right)\, dx\)
There is no partial fraction decomposition yet because long division must be done first.
\(\displaystyle \displaystyle \int \left( \frac{A}{x - 9} + \frac{B}{(x^2-1)^2} + \frac{C}{(x^2 + 9)^2} \right)\, dx\)
\(\displaystyle \displaystyle \int \left( \frac{A}{x - 9} + \frac{B}{x+1} + \frac{C}{(x+1)^2} + \frac{D}{x-1} + \frac{E}{(x-1)^2} + \frac{Fx+G}{x^2 + 9} + \frac{Hx+I}{(x^2 + 9)^2} \right)\, dx\)
There is no partial fraction decomposition yet because there is cancellation.
\(\displaystyle \displaystyle \int \left( \frac{A}{x - 9} + \frac{B}{x+1} + \frac{C}{(x+1)^2} + \frac{D}{x-1} + \frac{E}{(x-1)^2} + \frac{Fx+G}{x^2 + 9} \right)\, dx\)
There is no partial fraction decomposition because the denominator does not factor.
\(\displaystyle \displaystyle \int \left( \frac{A}{x - 9} + \frac{Bx+C}{(x^2-1)^2} + \frac{Dx+E}{(x^2 + 9)^2} \right)\, dx\)

5.

Consider the indefinite integral \(\displaystyle \int \frac{6 x^3+3 x^2 - 144 x - 65}{x^2-25}\, dx\)

Then the integrand decomposes into the form

\begin{equation*} a x + b + \frac{c}{x-5} +\frac{d}{x+5} \end{equation*}

where

\(a\) =

\(b\) =

\(c\) =

\(d\) =

Integrating term by term, we obtain that

\(\displaystyle \int \frac{6 x^3+3 x^2 - 144 x - 65}{x^2-25}\, dx =\)

\(+C\)

6.

Evaluate the integral

\begin{equation*} \int {\frac{-2}{(x+a)(x+b)}}\, dx \end{equation*}

for the cases where \(a=b\) and where \(a \ne b.\)

Note: For the case where \(a=b,\) use only \(a\) in your answer. Also, use an upper-case "C" for the constant of integration.

If \(a=b:\)

If \(a \ne b:\)

7.

Consider the integral

\begin{equation*} \int \frac{x^{21}-7 x^{14}+5 x^7-39}{\left(x^3-5 x^2+4 x\right)^3 \left(x^4-256\right)^2} \,dx \end{equation*}

Enter a T or an F in each answer space below to indicate whether or not a term of the given type occurs in the general form of the complete partial fractions decomposition of the integrand. \(A_1, A_2, A_3\dots\) and \(B_1, B_2, B_3,\dots\) denote constants.

You must get all of the answers correct to receive credit.

8.

Calculate the integral below by partial fractions and by using the indicated substitution. Be sure that you can show how the results you obtain are the same.

\begin{equation*} \int \frac{2x}{x^{2}-9}\,dx \end{equation*}

First, rewrite this with partial fractions:

\(\int \frac{2x}{x^{2}-9}\,dx = \int\) \(dx + \int\) \(dx =\) + \(+ C\text{.}\)

(Note that you should not include the \(+C\) in your entered answer, as it has been provided at the end of the expression.)

Next, use the substitution \(w = x^2 - 9\) to find the integral:

\(\int \frac{2x}{x^{2}-9}\,dx = \int\) \(dw =\) \(+ C =\) \(+ C\text{.}\)

(For the second answer blank, give your antiderivative in terms of the variable \(w\text{.}\) Again, note that you should not include the \(+C\) in your answer.)

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Section 9.2 Integrating Rational Functions

Subsection 9.2.1 Preview Activity

Question 9.2.1.

Question 9.2.2.

Question 9.2.3.

Question 9.2.4.

Exercises 9.2.2 Exercises

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