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Active Calculus

Section 9.3 Integrating Rational Functions

Subsection 9.3.1 Preview Activity

Question 9.3.1.

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.
11+x2dx
  • Basic Integral
  • u-Substitution
  • Integration by Parts
  • A combination of techniques
  • None of the above
x1+x2dx
  • Basic Integral
  • u-Substitution
  • Integration by Parts
  • A combination of techniques
  • None of the above
2x+31+x2dx
  • Basic Integral
  • u-Substitution
  • Integration by Parts
  • A combination of techniques
  • None of the above
ex1+(ex)2dx
  • Basic Integral
  • u-Substitution
  • Integration by Parts
  • A combination of techniques
  • None of the above
1x21dx
  • Basic Integral
  • u-Substitution
  • Integration by Parts
  • A combination of techniques
  • None of the above
Hint.
Note that 2x+31+x2 can be rewritten as 2x1+x2+31+x2.

Question 9.3.2.

First, note that we can use algebra to combine 8x+3+2x3 into one fraction by finding a common denominator.
8x+3+2x3=8x+3x3x3+2x3x+3x+3=306xx29
Also note that we know how to integrate
8x+3dx= and
2x3dx=
Which means we can integrate both sides of 8x+3+2x3=306xx29 to get
306xx29dx=
Hint.
Remember that you really should use absolute value instead of parentheses when you integrate to get ln. Sometimes WeBWorK doesn’t care about the difference, but sometimes it does. For example, 1x+11dx=ln|x+11|+C, not ln(x+11)+C.

Question 9.3.3.

Find the quotient and remainder using long division for
x5x4+x3x2+3x12x1.
The quotient is
The remainder is

Question 9.3.4.

Perform the indicated division and write the quotient and remainder in the provided blanks.
3x2x6x1
Answer: + /(x1)

Exercises 9.3.2 Exercises

1.

Which of the following is the correct form of the partial fraction decomposition of x1x3+x2? You may select more than one answer.
  1. Ax+Bx+Cx+Dx2+Ex+1
  2. Ax+Bx2+Cx+1
  3. Ax+Bx+Cx+Dx2+Ex+Fx+1
  4. Ax+Bx+Cx+Dx2+Ex+Fx+1

2.

Which of the following is the correct form of the partial fraction decomposition of x1x3+x? You may select more than one answer.
  1. Ax+Bx+Cx+Dx2+1
  2. Ax+Bx2+1
  3. Ax+Bx+Cx2+1
  4. Ax+Bx+Cx2+1

3.

What is the correct form of the partial fraction decomposition for the following integral?
x10x3+4x245xdx
  • A. (Ax+Bx5+Cx+9)dx
  • B. (Ax+Bx+Cx10+Dx+Ex+5)dx
  • C. There is no partial fraction decomposition because the denominator does not factor.
  • D. (Ax10+Bx9+Cx+5)dx
  • E. There is no partial fraction decomposition yet because there is cancellation.
  • F. There is no partial fraction decomposition yet because long division must be done first.
  • G. (Ax+Bx10+Cx+5)dx
  • H. (Ax+Bx+Cx+Dx5+Ex+Fx+9)dx

4.

What is the correct form of the partial fraction decomposition for the following integral?
10(x8+6)(x7)(x21)2(x2+7)2dx
  • A. There is no partial fraction decomposition because the denominator does not factor.
  • B. (Ax7+Bx+1+C(x+1)2+Dx1+E(x1)2+Fx+Gx2+7)dx
  • C. (Ax7+Bx+C(x21)2+Dx+E(x2+7)2)dx
  • D. (Ax7+Bx+Cx21+Cx+D(x21)2+Ex+Fx2+7+Gx+H(x2+7)2)dx
  • E. (Ax7+B(x21)2+C(x2+7)2)dx
  • F. There is no partial fraction decomposition yet because long division must be done first.
  • G. There is no partial fraction decomposition yet because there is cancellation.

5.

Consider the indefinite integral 6x3+2x249x21x29dx
Then the integrand decomposes into the form
ax+b+cx3+dx+3
where
a =
b =
c =
d =
Integrating term by term, we obtain that
6x3+2x249x21x29dx=
+C

6.

Evaluate the integral
10(x+a)(x+b)dx
for the cases where a=b and where ab.
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 ab:

7.

Consider the integral
x213x14+6x716(x36x2+5x)3(x4625)2dx
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. A1,A2,A3 and B1,B2,B3, 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.
2xx24dx
First, rewrite this with partial fractions:
2xx24dx= dx+ dx= + +C.
(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=x24 to find the integral:
2xx24dx= dw= +C= +C.
(For the second answer blank, give your antiderivative in terms of the variable w. Again, note that you should not include the +C in your answer.)
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