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Calculus Volume 2

3.4 Partial Fractions

Calculus Volume 23.4 Partial Fractions
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  1. Preface
  2. 1 Integration
    1. Introduction
    2. 1.1 Approximating Areas
    3. 1.2 The Definite Integral
    4. 1.3 The Fundamental Theorem of Calculus
    5. 1.4 Integration Formulas and the Net Change Theorem
    6. 1.5 Substitution
    7. 1.6 Integrals Involving Exponential and Logarithmic Functions
    8. 1.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  3. 2 Applications of Integration
    1. Introduction
    2. 2.1 Areas between Curves
    3. 2.2 Determining Volumes by Slicing
    4. 2.3 Volumes of Revolution: Cylindrical Shells
    5. 2.4 Arc Length of a Curve and Surface Area
    6. 2.5 Physical Applications
    7. 2.6 Moments and Centers of Mass
    8. 2.7 Integrals, Exponential Functions, and Logarithms
    9. 2.8 Exponential Growth and Decay
    10. 2.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  4. 3 Techniques of Integration
    1. Introduction
    2. 3.1 Integration by Parts
    3. 3.2 Trigonometric Integrals
    4. 3.3 Trigonometric Substitution
    5. 3.4 Partial Fractions
    6. 3.5 Other Strategies for Integration
    7. 3.6 Numerical Integration
    8. 3.7 Improper Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  5. 4 Introduction to Differential Equations
    1. Introduction
    2. 4.1 Basics of Differential Equations
    3. 4.2 Direction Fields and Numerical Methods
    4. 4.3 Separable Equations
    5. 4.4 The Logistic Equation
    6. 4.5 First-order Linear Equations
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  6. 5 Sequences and Series
    1. Introduction
    2. 5.1 Sequences
    3. 5.2 Infinite Series
    4. 5.3 The Divergence and Integral Tests
    5. 5.4 Comparison Tests
    6. 5.5 Alternating Series
    7. 5.6 Ratio and Root Tests
    8. Key Terms
    9. Key Equations
    10. Key Concepts
    11. Chapter Review Exercises
  7. 6 Power Series
    1. Introduction
    2. 6.1 Power Series and Functions
    3. 6.2 Properties of Power Series
    4. 6.3 Taylor and Maclaurin Series
    5. 6.4 Working with Taylor Series
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  8. 7 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 7.1 Parametric Equations
    3. 7.2 Calculus of Parametric Curves
    4. 7.3 Polar Coordinates
    5. 7.4 Area and Arc Length in Polar Coordinates
    6. 7.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 3.4.1. Integrate a rational function using the method of partial fractions.
  • 3.4.2. Recognize simple linear factors in a rational function.
  • 3.4.3. Recognize repeated linear factors in a rational function.
  • 3.4.4. Recognize quadratic factors in a rational function.

We have seen some techniques that allow us to integrate specific rational functions. For example, we know that

duu=ln|u|+Candduu2+a2=1atan−1(ua)+C.duu=ln|u|+Candduu2+a2=1atan−1(ua)+C.

However, we do not yet have a technique that allows us to tackle arbitrary quotients of this type. Thus, it is not immediately obvious how to go about evaluating 3xx2x2dx.3xx2x2dx. However, we know from material previously developed that

(1x+1+2x2)dx=ln|x+1|+2ln|x2|+C.(1x+1+2x2)dx=ln|x+1|+2ln|x2|+C.

In fact, by getting a common denominator, we see that

1x+1+2x2=3xx2x2.1x+1+2x2=3xx2x2.

Consequently,

3xx2x2dx=(1x+1+2x2)dx.3xx2x2dx=(1x+1+2x2)dx.

In this section, we examine the method of partial fraction decomposition, which allows us to decompose rational functions into sums of simpler, more easily integrated rational functions. Using this method, we can rewrite an expression such as: 3xx2x23xx2x2 as an expression such as 1x+1+2x2.1x+1+2x2.

The key to the method of partial fraction decomposition is being able to anticipate the form that the decomposition of a rational function will take. As we shall see, this form is both predictable and highly dependent on the factorization of the denominator of the rational function. It is also extremely important to keep in mind that partial fraction decomposition can be applied to a rational function P(x)Q(x)P(x)Q(x) only if deg(P(x))<deg(Q(x)).deg(P(x))<deg(Q(x)). In the case when deg(P(x))deg(Q(x)),deg(P(x))deg(Q(x)), we must first perform long division to rewrite the quotient P(x)Q(x)P(x)Q(x) in the form A(x)+R(x)Q(x),A(x)+R(x)Q(x), where deg(R(x))<deg(Q(x)).deg(R(x))<deg(Q(x)). We then do a partial fraction decomposition on R(x)Q(x).R(x)Q(x). The following example, although not requiring partial fraction decomposition, illustrates our approach to integrals of rational functions of the form P(x)Q(x)dx,P(x)Q(x)dx, where deg(P(x))deg(Q(x)).deg(P(x))deg(Q(x)).

Example 3.28

Integrating P(x)Q(x)dx,P(x)Q(x)dx, where deg(P(x))deg(Q(x))deg(P(x))deg(Q(x))

Evaluate x2+3x+5x+1dx.x2+3x+5x+1dx.

Solution

Since deg(x2+3x+5)deg(x+1),deg(x2+3x+5)deg(x+1), we perform long division to obtain

x2+3x+5x+1=x+2+3x+1.x2+3x+5x+1=x+2+3x+1.

Thus,

x2+3x+5x+1dx=(x+2+3x+1)dx=12x2+2x+3ln|x+1|+C.x2+3x+5x+1dx=(x+2+3x+1)dx=12x2+2x+3ln|x+1|+C.

Media

Visit this website for a review of long division of polynomials.

Checkpoint 3.17

Evaluate x3x+2dx.x3x+2dx.

To integrate P(x)Q(x)dx,P(x)Q(x)dx, where deg(P(x))<deg(Q(x)),deg(P(x))<deg(Q(x)), we must begin by factoring Q(x).Q(x).

Nonrepeated Linear Factors

If Q(x)Q(x) can be factored as (a1x+b1)(a2x+b2)(anx+bn),(a1x+b1)(a2x+b2)(anx+bn), where each linear factor is distinct, then it is possible to find constants A1,A2,…AnA1,A2,…An satisfying

P(x)Q(x)=A1a1x+b1+A2a2x+b2++Ananx+bn.P(x)Q(x)=A1a1x+b1+A2a2x+b2++Ananx+bn.

The proof that such constants exist is beyond the scope of this course.

In this next example, we see how to use partial fractions to integrate a rational function of this type.

Example 3.29

Partial Fractions with Nonrepeated Linear Factors

Evaluate 3x+2x3x22xdx.3x+2x3x22xdx.

Solution

Since deg(3x+2)<deg(x3x22x),deg(3x+2)<deg(x3x22x), we begin by factoring the denominator of 3x+2x3x22x.3x+2x3x22x. We can see that x3x22x=x(x2)(x+1).x3x22x=x(x2)(x+1). Thus, there are constants A,A, B,B, and CC satisfying

3x+2x(x2)(x+1)=Ax+Bx2+Cx+1.3x+2x(x2)(x+1)=Ax+Bx2+Cx+1.

We must now find these constants. To do so, we begin by getting a common denominator on the right. Thus,

3x+2x(x2)(x+1)=A(x2)(x+1)+Bx(x+1)+Cx(x2)x(x2)(x+1).3x+2x(x2)(x+1)=A(x2)(x+1)+Bx(x+1)+Cx(x2)x(x2)(x+1).

Now, we set the numerators equal to each other, obtaining

3x+2=A(x2)(x+1)+Bx(x+1)+Cx(x2).3x+2=A(x2)(x+1)+Bx(x+1)+Cx(x2).
3.8

There are two different strategies for finding the coefficients A,A, B,B, and C.C. We refer to these as the method of equating coefficients and the method of strategic substitution.

Rule: Method of Equating Coefficients

Rewrite Equation 3.8 in the form

3x+2=(A+B+C)x2+(A+B2C)x+(−2A).3x+2=(A+B+C)x2+(A+B2C)x+(−2A).

Equating coefficients produces the system of equations

A+B+C=0A+B2C=32A=2.A+B+C=0A+B2C=32A=2.

To solve this system, we first observe that −2A=2A=−1.−2A=2A=−1. Substituting this value into the first two equations gives us the system

B+C=1B2C=2.B+C=1B2C=2.

Multiplying the second equation by −1−1 and adding the resulting equation to the first produces

−3C=1,−3C=1,

which in turn implies that C=13.C=13. Substituting this value into the equation B+C=1B+C=1 yields B=43.B=43. Thus, solving these equations yields A=−1,A=−1, B=43,B=43, and C=13.C=13.

It is important to note that the system produced by this method is consistent if and only if we have set up the decomposition correctly. If the system is inconsistent, there is an error in our decomposition.

Rule: Method of Strategic Substitution

The method of strategic substitution is based on the assumption that we have set up the decomposition correctly. If the decomposition is set up correctly, then there must be values of A,A, B,B, and CC that satisfy Equation 3.8 for all values of x.x. That is, this equation must be true for any value of xx we care to substitute into it. Therefore, by choosing values of xx carefully and substituting them into the equation, we may find A,A, B,B, and CC easily. For example, if we substitute x=0,x=0, the equation reduces to 2=A(−2)(1).2=A(−2)(1). Solving for AA yields A=−1.A=−1. Next, by substituting x=2,x=2, the equation reduces to 8=B(2)(3),8=B(2)(3), or equivalently B=4/3.B=4/3. Last, we substitute x=−1x=−1 into the equation and obtain −1=C(−1)(−3).−1=C(−1)(−3). Solving, we have C=13.C=13.

It is important to keep in mind that if we attempt to use this method with a decomposition that has not been set up correctly, we are still able to find values for the constants, but these constants are meaningless. If we do opt to use the method of strategic substitution, then it is a good idea to check the result by recombining the terms algebraically.

Now that we have the values of A,A, B,B, and C,C, we rewrite the original integral:

3x+2x3x22xdx=(1x+43·1(x2)13·1(x+1))dx.3x+2x3x22xdx=(1x+43·1(x2)13·1(x+1))dx.

Evaluating the integral gives us

3x+2x3x22xdx=ln|x|+43ln|x2|13ln|x+1|+C.3x+2x3x22xdx=ln|x|+43ln|x2|13ln|x+1|+C.

In the next example, we integrate a rational function in which the degree of the numerator is not less than the degree of the denominator.

Example 3.30

Dividing before Applying Partial Fractions

Evaluate x2+3x+1x24dx.x2+3x+1x24dx.

Solution

Since degree(x2+3x+1)degree(x24),degree(x2+3x+1)degree(x24), we must perform long division of polynomials. This results in

x2+3x+1x24=1+3x+5x24.x2+3x+1x24=1+3x+5x24.

Next, we perform partial fraction decomposition on 3x+5x24=3x+5(x+2)(x2).3x+5x24=3x+5(x+2)(x2). We have

3x+5(x2)(x+2)=Ax2+Bx+2.3x+5(x2)(x+2)=Ax2+Bx+2.

Thus,

3x+5=A(x+2)+B(x2).3x+5=A(x+2)+B(x2).

Solving for AA and BB using either method, we obtain A=11/4A=11/4 and B=1/4.B=1/4.

Rewriting the original integral, we have

x2+3x+1x24dx=(1+114·1x2+14·1x+2)dx.x2+3x+1x24dx=(1+114·1x2+14·1x+2)dx.

Evaluating the integral produces

x2+3x+1x24dx=x+114ln|x2|+14ln|x+2|+C.x2+3x+1x24dx=x+114ln|x2|+14ln|x+2|+C.

As we see in the next example, it may be possible to apply the technique of partial fraction decomposition to a nonrational function. The trick is to convert the nonrational function to a rational function through a substitution.

Example 3.31

Applying Partial Fractions after a Substitution

Evaluate cosxsin2xsinxdx.cosxsin2xsinxdx.

Solution

Let’s begin by letting u=sinx.u=sinx. Consequently, du=cosxdx.du=cosxdx. After making these substitutions, we have

cosxsin2xsinxdx=duu2u=duu(u1).cosxsin2xsinxdx=duu2u=duu(u1).

Applying partial fraction decomposition to 1/u(u1)1/u(u1) gives 1u(u1)=1u+1u1.1u(u1)=1u+1u1.

Thus,

cosxsin2xsinxdx=ln|u|+ln|u1|+C=ln|sinx|+ln|sinx1|+C.cosxsin2xsinxdx=ln|u|+ln|u1|+C=ln|sinx|+ln|sinx1|+C.
Checkpoint 3.18

Evaluate x+1(x+3)(x2)dx.x+1(x+3)(x2)dx.

Repeated Linear Factors

For some applications, we need to integrate rational expressions that have denominators with repeated linear factors—that is, rational functions with at least one factor of the form (ax+b)n,(ax+b)n, where nn is a positive integer greater than or equal to 2.2. If the denominator contains the repeated linear factor (ax+b)n,(ax+b)n, then the decomposition must contain

A1ax+b+A2(ax+b)2++An(ax+b)n.A1ax+b+A2(ax+b)2++An(ax+b)n.

As we see in our next example, the basic technique used for solving for the coefficients is the same, but it requires more algebra to determine the numerators of the partial fractions.

Example 3.32

Partial Fractions with Repeated Linear Factors

Evaluate x2(2x1)2(x1)dx.x2(2x1)2(x1)dx.

Solution

We have degree(x2)<degree((2x1)2(x1)),degree(x2)<degree((2x1)2(x1)), so we can proceed with the decomposition. Since (2x1)2(2x1)2 is a repeated linear factor, include A2x1+B(2x1)2A2x1+B(2x1)2 in the decomposition. Thus,

x2(2x1)2(x1)=A2x1+B(2x1)2+Cx1.x2(2x1)2(x1)=A2x1+B(2x1)2+Cx1.

After getting a common denominator and equating the numerators, we have

x2=A(2x1)(x1)+B(x1)+C(2x1)2.x2=A(2x1)(x1)+B(x1)+C(2x1)2.
3.9

We then use the method of equating coefficients to find the values of A,A, B,B, and C.C.

x2=(2A+4C)x2+(−3A+B4C)x+(AB+C).x2=(2A+4C)x2+(−3A+B4C)x+(AB+C).

Equating coefficients yields 2A+4C=0,2A+4C=0, −3A+B4C=1,−3A+B4C=1, and AB+C=−2.AB+C=−2. Solving this system yields A=2,A=2, B=3,B=3, and C=−1.C=−1.

Alternatively, we can use the method of strategic substitution. In this case, substituting x=1x=1 and x=1/2x=1/2 into Equation 3.9 easily produces the values B=3B=3 and C=−1.C=−1. At this point, it may seem that we have run out of good choices for x,x, however, since we already have values for BB and C,C, we can substitute in these values and choose any value for xx not previously used. The value x=0x=0 is a good option. In this case, we obtain the equation −2=A(−1)(−1)+3(−1)+(−1)(−1)2−2=A(−1)(−1)+3(−1)+(−1)(−1)2 or, equivalently, A=2.A=2.

Now that we have the values for A,A, B,B, and C,C, we rewrite the original integral and evaluate it:

x2(2x1)2(x1)dx=(22x1+3(2x1)21x1)dx=ln|2x1|32(2x1)ln|x1|+C.x2(2x1)2(x1)dx=(22x1+3(2x1)21x1)dx=ln|2x1|32(2x1)ln|x1|+C.
Checkpoint 3.19

Set up the partial fraction decomposition for x+2(x+3)3(x4)2dx.x+2(x+3)3(x4)2dx. (Do not solve for the coefficients or complete the integration.)

The General Method

Now that we are beginning to get the idea of how the technique of partial fraction decomposition works, let’s outline the basic method in the following problem-solving strategy.

Problem-Solving Strategy: Partial Fraction Decomposition

To decompose the rational function P(x)/Q(x),P(x)/Q(x), use the following steps:

  1. Make sure that degree(P(x))<degree(Q(x)).degree(P(x))<degree(Q(x)). If not, perform long division of polynomials.
  2. Factor Q(x)Q(x) into the product of linear and irreducible quadratic factors. An irreducible quadratic is a quadratic that has no real zeros.
  3. Assuming that deg(P(x))<deg(Q(x)),deg(P(x))<deg(Q(x)), the factors of Q(x)Q(x) determine the form of the decomposition of P(x)/Q(x).P(x)/Q(x).
    1. If Q(x)Q(x) can be factored as (a1x+b1)(a2x+b2)(anx+bn),(a1x+b1)(a2x+b2)(anx+bn), where each linear factor is distinct, then it is possible to find constants A1,A2,...AnA1,A2,...An satisfying
      P(x)Q(x)=A1a1x+b1+A2a2x+b2++Ananx+bn.P(x)Q(x)=A1a1x+b1+A2a2x+b2++Ananx+bn.
    2. If Q(x)Q(x) contains the repeated linear factor (ax+b)n,(ax+b)n, then the decomposition must contain
      A1ax+b+A2(ax+b)2++An(ax+b)n.A1ax+b+A2(ax+b)2++An(ax+b)n.
    3. For each irreducible quadratic factor ax2+bx+cax2+bx+c that Q(x)Q(x) contains, the decomposition must include
      Ax+Bax2+bx+c.Ax+Bax2+bx+c.
    4. For each repeated irreducible quadratic factor (ax2+bx+c)n,(ax2+bx+c)n, the decomposition must include
      A1x+B1ax2+bx+c+A2x+B2(ax2+bx+c)2++Anx+Bn(ax2+bx+c)n.A1x+B1ax2+bx+c+A2x+B2(ax2+bx+c)2++Anx+Bn(ax2+bx+c)n.
    5. After the appropriate decomposition is determined, solve for the constants.
    6. Last, rewrite the integral in its decomposed form and evaluate it using previously developed techniques or integration formulas.

Simple Quadratic Factors

Now let’s look at integrating a rational expression in which the denominator contains an irreducible quadratic factor. Recall that the quadratic ax2+bx+cax2+bx+c is irreducible if ax2+bx+c=0ax2+bx+c=0 has no real zeros—that is, if b24ac<0.b24ac<0.

Example 3.33

Rational Expressions with an Irreducible Quadratic Factor

Evaluate 2x3x3+xdx.2x3x3+xdx.

Solution

Since deg(2x3)<deg(x3+x),deg(2x3)<deg(x3+x), factor the denominator and proceed with partial fraction decomposition. Since x3+x=x(x2+1)x3+x=x(x2+1) contains the irreducible quadratic factor x2+1,x2+1, include Ax+Bx2+1Ax+Bx2+1 as part of the decomposition, along with CxCx for the linear term x.x. Thus, the decomposition has the form

2x3x(x2+1)=Ax+Bx2+1+Cx.2x3x(x2+1)=Ax+Bx2+1+Cx.

After getting a common denominator and equating the numerators, we obtain the equation

2x3=(Ax+B)x+C(x2+1).2x3=(Ax+B)x+C(x2+1).

Solving for A,B,A,B, and C,C, we get A=3,A=3, B=2,B=2, and C=−3.C=−3.

Thus,

2x3x3+x=3x+2x2+13x.2x3x3+x=3x+2x2+13x.

Substituting back into the integral, we obtain

2x3x3+xdx=(3x+2x2+13x)dx=3xx2+1dx+21x2+1dx31xdxSplit up the integral.=32ln|x2+1|+2tan−1x3ln|x|+C.Evaluate each integral.2x3x3+xdx=(3x+2x2+13x)dx=3xx2+1dx+21x2+1dx31xdxSplit up the integral.=32ln|x2+1|+2tan−1x3ln|x|+C.Evaluate each integral.

Note: We may rewrite ln|x2+1|=ln(x2+1),ln|x2+1|=ln(x2+1), if we wish to do so, since x2+1>0.x2+1>0.

Example 3.34

Partial Fractions with an Irreducible Quadratic Factor

Evaluate dxx38.dxx38.

Solution

We can start by factoring x38=(x2)(x2+2x+4).x38=(x2)(x2+2x+4). We see that the quadratic factor x2+2x+4x2+2x+4 is irreducible since 224(1)(4)=−12<0.224(1)(4)=−12<0. Using the decomposition described in the problem-solving strategy, we get

1(x2)(x2+2x+4)=Ax2+Bx+Cx2+2x+4.1(x2)(x2+2x+4)=Ax2+Bx+Cx2+2x+4.

After obtaining a common denominator and equating the numerators, this becomes

1=A(x2+2x+4)+(Bx+C)(x2).1=A(x2+2x+4)+(Bx+C)(x2).

Applying either method, we get A=112,B=112,andC=13.A=112,B=112,andC=13.

Rewriting dxx38,dxx38, we have

dxx38=1121x2dx112x+4x2+2x+4dx.dxx38=1121x2dx112x+4x2+2x+4dx.

We can see that

1x2dx=ln|x2|+C,1x2dx=ln|x2|+C, but x+4x2+2x+4dxx+4x2+2x+4dx requires a bit more effort. Let’s begin by completing the square on x2+2x+4x2+2x+4 to obtain

x2+2x+4=(x+1)2+3.x2+2x+4=(x+1)2+3.

By letting u=x+1u=x+1 and consequently du=dx,du=dx, we see that

x+4x2+2x+4dx=x+4(x+1)2+3dxComplete the square on thedenominator.=u+3u2+3duSubstituteu=x+1,x=u1,anddu=dx.=uu2+3du+3u2+3duSplit the numerator apart.=12ln|u2+3|+33tan−1u3+CEvaluate each integral.=12ln|x2+2x+4|+3tan−1(x+13)+C.Rewrite in terms ofxandsimplify.x+4x2+2x+4dx=x+4(x+1)2+3dxComplete the square on thedenominator.=u+3u2+3duSubstituteu=x+1,x=u1,anddu=dx.=uu2+3du+3u2+3duSplit the numerator apart.=12ln|u2+3|+33tan−1u3+CEvaluate each integral.=12ln|x2+2x+4|+3tan−1(x+13)+C.Rewrite in terms ofxandsimplify.

Substituting back into the original integral and simplifying gives

dxx38=112ln|x2|124ln|x2+2x+4|312tan−1(x+13)+C.dxx38=112ln|x2|124ln|x2+2x+4|312tan−1(x+13)+C.

Here again, we can drop the absolute value if we wish to do so, since x2+2x+4>0x2+2x+4>0 for all x.x.

Example 3.35

Finding a Volume

Find the volume of the solid of revolution obtained by revolving the region enclosed by the graph of f(x)=x2(x2+1)2f(x)=x2(x2+1)2 and the x-axis over the interval [0,1][0,1] about the y-axis.

Solution

Let’s begin by sketching the region to be revolved (see Figure 3.11). From the sketch, we see that the shell method is a good choice for solving this problem.

This figure is the graph of the function f(x) = x^2/(x^2+1)^2. It is a curve above the x-axis. It is decreasing in the second quadrant, intersects at the origin, and increases in the first quadrant. Between x = 0 and x = 1, there is shaded area under the curve.
Figure 3.11 We can use the shell method to find the volume of revolution obtained by revolving the region shown about the y-axis.

The volume is given by

V=2π01x·x2(x2+1)2dx=2π01x3(x2+1)2dx.V=2π01x·x2(x2+1)2dx=2π01x3(x2+1)2dx.

Since deg((x2+1)2)=4>3=deg(x3),deg((x2+1)2)=4>3=deg(x3), we can proceed with partial fraction decomposition. Note that (x2+1)2(x2+1)2 is a repeated irreducible quadratic. Using the decomposition described in the problem-solving strategy, we get

x3(x2+1)2=Ax+Bx2+1+Cx+D(x2+1)2.x3(x2+1)2=Ax+Bx2+1+Cx+D(x2+1)2.

Finding a common denominator and equating the numerators gives

x3=(Ax+B)(x2+1)+Cx+D.x3=(Ax+B)(x2+1)+Cx+D.

Solving, we obtain A=1,A=1, B=0,B=0, C=−1,C=−1, and D=0.D=0. Substituting back into the integral, we have

V=2π01x3(x2+1)2dx=2π01(xx2+1x(x2+1)2)dx=2π(12ln(x2+1)+12·1x2+1)|01=π(ln212).V=2π01x3(x2+1)2dx=2π01(xx2+1x(x2+1)2)dx=2π(12ln(x2+1)+12·1x2+1)|01=π(ln212).
Checkpoint 3.20

Set up the partial fraction decomposition for x2+3x+1(x+2)(x3)2(x2+4)2dx.x2+3x+1(x+2)(x3)2(x2+4)2dx.

Section 3.4 Exercises

Express the rational function as a sum or difference of two simpler rational expressions.

182.

1(x3)(x2)1(x3)(x2)

183.

x2+1x(x+1)(x+2)x2+1x(x+1)(x+2)

184.

1x3x1x3x

185.

3x+1x23x+1x2

186.

3x2x2+13x2x2+1 (Hint: Use long division first.)

187.

2x4x22x2x4x22x

188.

1(x1)(x2+1)1(x1)(x2+1)

189.

1x2(x1)1x2(x1)

190.

xx24xx24

191.

1x(x1)(x2)(x3)1x(x1)(x2)(x3)

192.

1x41=1(x+1)(x1)(x2+1)1x41=1(x+1)(x1)(x2+1)

193.

3x2x31=3x2(x1)(x2+x+1)3x2x31=3x2(x1)(x2+x+1)

194.

2x(x+2)22x(x+2)2

195.

3x4+x3+20x2+3x+31(x+1)(x2+4)23x4+x3+20x2+3x+31(x+1)(x2+4)2

Use the method of partial fractions to evaluate each of the following integrals.

196.

dx(x3)(x2)dx(x3)(x2)

197.

3xx2+2x8dx3xx2+2x8dx

198.

dxx3xdxx3x

199.

xx24dxxx24dx

200.

dxx(x1)(x2)(x3)dxx(x1)(x2)(x3)

201.

2x2+4x+22x2+2x+10dx2x2+4x+22x2+2x+10dx

202.

dxx25x+6dxx25x+6

203.

2xx2+xdx2xx2+xdx

204.

2x2x6dx2x2x6dx

205.

dxx32x24x+8dxx32x24x+8

206.

dxx410x2+9dxx410x2+9

Evaluate the following integrals, which have irreducible quadratic factors.

207.

2(x4)(x2+2x+6)dx2(x4)(x2+2x+6)dx

208.

x2x3x2+4x4dxx2x3x2+4x4dx

209.

x3+6x2+3x+6x3+2x2dxx3+6x2+3x+6x3+2x2dx

210.

x(x1)(x2+2x+2)2dxx(x1)(x2+2x+2)2dx

Use the method of partial fractions to evaluate the following integrals.

211.

3x+4(x2+4)(3x)dx3x+4(x2+4)(3x)dx

212.

2(x+2)2(2x)dx2(x+2)2(2x)dx

213.

3x+4x32x4dx3x+4x32x4dx (Hint: Use the rational root theorem.)

Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals.

214.

01ex36e2xdx01ex36e2xdx (Give the exact answer and the decimal equivalent. Round to five decimal places.)

215.

exdxe2xexdxexdxe2xexdx

216.

sinxdx1cos2xsinxdx1cos2x

217.

sinxcos2x+cosx6dxsinxcos2x+cosx6dx

218.

1x1+xdx1x1+xdx

219.

dt(etet)2dt(etet)2

220.

1+ex1exdx1+ex1exdx

221.

dx1+x+1dx1+x+1

222.

dxx+x4dxx+x4

223.

cosxsinx(1sinx)dxcosxsinx(1sinx)dx

224.

ex(e2x4)2dxex(e2x4)2dx

225.

121x24x2dx121x24x2dx

226.

12+exdx12+exdx

227.

11+exdx11+exdx

Use the given substitution to convert the integral to an integral of a rational function, then evaluate.

228.

1tt3dtt=x31tt3dtt=x3

229.

1x+x3dx;x=u61x+x3dx;x=u6

230.

Graph the curve y=x1+xy=x1+x over the interval [0,5].[0,5]. Then, find the area of the region bounded by the curve, the x-axis, and the line x=4.x=4.

This figure is a graph of the function y = x/(1 + x). The graph is only in the first quadrant. It begins at the origin and increases into the first quadrant. The curve stops at x = 5.
231.

Find the volume of the solid generated when the region bounded by y=1/x(3x),y=1/x(3x), y=0,y=0, x=1,x=1, and x=2x=2 is revolved about the x-axis.

232.

The velocity of a particle moving along a line is a function of time given by v(t)=88t2t2+1.v(t)=88t2t2+1. Find the distance that the particle has traveled after t=5t=5 sec.

Solve the initial-value problem for x as a function of t.

233.

(t27t+12)dxdt=1,(t>4,x(5)=0)(t27t+12)dxdt=1,(t>4,x(5)=0)

234.

(t+5)dxdt=x2+1,t>5,x(1)=tan1(t+5)dxdt=x2+1,t>5,x(1)=tan1

235.

(2t32t2+t1)dxdt=3,x(2)=0(2t32t2+t1)dxdt=3,x(2)=0

236.

Find the x-coordinate of the centroid of the area bounded by

y(x29)=1,y(x29)=1, y=0,x=4,andx=5.y=0,x=4,andx=5. (Round the answer to two decimal places.)

237.

Find the volume generated by revolving the area bounded by y=1x3+7x2+6x,x=1,x=7,andy=0y=1x3+7x2+6x,x=1,x=7,andy=0 about the y-axis.

238.

Find the area bounded by y=x12x28x20,y=x12x28x20, y=0,x=2,andx=4.y=0,x=2,andx=4. (Round the answer to the nearest hundredth.)

239.

Evaluate the integral dxx3+1.dxx3+1.

For the following problems, use the substitutions tan(x2)=t,tan(x2)=t, dx=21+t2dt,dx=21+t2dt, sinx=2t1+t2,sinx=2t1+t2, and cosx=1t21+t2.cosx=1t21+t2.

240.

dx35sinxdx35sinx

241.

Find the area under the curve y=11+sinxy=11+sinx between x=0x=0 and x=π.x=π. (Assume the dimensions are in inches.)

242.

Given tan(x2)=t,tan(x2)=t, derive the formulas dx=21+t2dt,dx=21+t2dt, sinx=2t1+t2,sinx=2t1+t2, and cosx=1t21+t2.cosx=1t21+t2.

243.

Evaluate x83xdx.x83xdx.

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