Integration Using Partial Fractions

8 Key excerpts on "Integration Using Partial Fractions"

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  Anton's Calculus

  Early Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2018(Publication Date)
  • Wiley

    (Publisher)

  This idea is illustrated in the following example. Example 5 Evaluate ∫ 3x 4 + 3x 3 − 5x 2 + x − 1 x 2 + x − 2 dx. 7.5 Integrating Rational Functions by Partial Fractions 443 Solution. The integrand is an improper rational function since the numerator has degree 4 and the denominator has degree 2. Thus, we first perform the long division 3x 2 + 1 x 2 + x − 2 3x 4 + 3x 3 − 5x 2 + x − 1 3x 4 + 3x 3 − 6x 2 x 2 + x − 1 x 2 + x − 2 1 It follows that the integrand can be expressed as 3x 4 + 3x 3 − 5x 2 + x − 1 x 2 + x − 2 = (3x 2 + 1) + 1 x 2 + x − 2 and hence ∫ 3x 4 + 3x 3 − 5x 2 + x − 1 x 2 + x − 2 dx = ∫ (3x 2 + 1) dx + ∫ dx x 2 + x − 2 The second integral on the right now involves a proper rational function and can thus be evaluated by a partial fraction decomposition. Using the result of Example 1 we obtain ∫ 3x 4 + 3x 3 − 5x 2 + x − 1 x 2 + x − 2 dx = x 3 + x + 1 3 ln | | | | x − 1 x + 2 | | | | + C CONCLUDING REMARKS There are some cases in which the method of partial fractions is inappropriate. For example, it would be inefficient to use partial fractions to perform the integration ∫ 3x 2 + 2 x 3 + 2x − 8 dx = ln |x 3 + 2x − 8| + C since the substitution u = x 3 + 2x − 8 is more direct. Similarly, the integration ∫ 2x − 1 x 2 + 1 dx = ∫ 2x x 2 + 1 dx − ∫ dx x 2 + 1 = ln(x 2 + 1) − tan −1 x + C requires only a little algebra since the integrand is already in partial fraction form. QUICK CHECK EXERCISES 7.5 (See page 445 for answers.) 1. A partial fraction is a rational function of the form or of the form . 2. (a) What is a proper rational function? (b) What condition must the degree of the numerator and the degree of the denominator of a rational function sat- isfy for the method of partial fractions to be applicable directly? (c) If the condition in part (b) is not satisfied, what must you do if you want to use partial fractions? 3. Suppose that the function f (x) = P(x) ∕ Q(x) is a proper rational function.

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  Sixth Form Pure Mathematics

  Volume 1

  • C. Plumpton, W. A. Tomkys(Authors)
  • 2014(Publication Date)
  • Pergamon

    (Publisher)

  CHAPTER X PARTIAL FRACTIONS AND THEIR APPLICATIONS. SOME FURTHER METHODS OF INTEGRATION 10.1 Partial Fractions Two or more algebraic fractions can be added to give a single com-pound fraction as in the following example. 3 2x4-1 1 _ 3x*+3x+2x*+3x 2 +x-x*-x 2 -x-l jc-fl * x 2 +1 x x(x+ l)(x 2 +1) _ 4x 3 +2x 2 +3;c-l x(x+l)(x*+l) • For many purposes in mathematical analysis it is necessary to perform the reverse process and to express a single compound fraction as the sum of two or more simple fractions. (The phrase compound fraction is used here to denote a fraction which has a denominator which can be expressed as the product of factors, and the phrase simple fraction to dencte a fraction which has a denominator which cannot be factorized.) A jomplete justification of the processes involved in obtaining partial fractions will not be given here. We make the following fundamental assumptions, using the term proper fraction to denote a fraction in which the degree of the numerator is less than the degree of the denominator. (a) Any given proper fraction whose denominator factorizes can be written as the sum of two or more proper fractions. (b)If f(x) = ao+axX+aix 2 +. . . + a n x n and g(x) = b 0 +b 1 x+b 2 x 2 +. . . + b n x n 9 then f(x) is said to be identically equal to g(x) when f(x) = g(x) for all values of x, and the relation is then written f(x) = g(x). 371 372 SIXTH FORM PURE MATHEMATICS /* can be proved that iff(x) = g(x then a r = b r for all values of r. oo oo This result is true for two infinite series also, i.e. if £ a^ = £ b^ for o o some range of values of x, then a r = b r . 4*-17 Examples, (i) Express ( X iA(2 x —3) i n P artia * fractions. 4x-17 A B Assume , ^ *x ~ ~TZ * 2~^3 w ^ e r e ^ anc * ^ a r e constants. Then 4x-17 = A(2x-3)+B(x+4). (1) Hence, equating coefficients of x and equating the absolute terms of (1), 2A + B = 4, -3A+4B = -17.

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  Calculus

  Single Variable

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  Using the result of Example 1, we obtain  3x 4 + 3x 3 − 5x 2 + x − 1 x 2 + x − 2 dx = x 3 + x + 1 3 ln     x − 1 x + 2     + C Concluding Remarks There are some cases in which the method of partial fractions is inappropriate. For example, it would be inefficient to use partial fractions to perform the integration  3x 2 + 2 x 3 + 2x − 8 dx = ln |x 3 + 2x − 8| + C since the substitution u = x 3 + 2x − 8 is more direct. Similarly, the integration  2x − 1 x 2 + 1 dx =  2x x 2 + 1 dx −  dx x 2 + 1 = ln(x 2 + 1) − tan −1 x + C requires only a little algebra since the integrand is already in partial fraction form. 7.5 | Quick Check Exercises (See page 437 for answers.) 1. A partial fraction is a rational function of the form or of the form . 2. a. What is a proper rational function? b. What condition must the degree of the numerator and the degree of the denominator of a rational function satisfy for the method of partial fractions to be applicable directly? c. If the condition in part (b) is not satisfied, what must you do if you want to use partial fractions? 3. Suppose that the function f (x) = P(x)/Q(x) is a proper rational function. a. For each factor of Q(x) of the form (ax + b) m , the partial fraction decomposition of f contains the following sum of m partial fractions: b. For each factor of Q(x) of the form (ax 2 + bx + c) m , where ax 2 + bx + c is an irreducible quadratic, the partial frac- tion decomposition of f contains the following sum of m partial fractions: 4. Complete the partial fraction decomposition. a. −3 (x + 1)(2x − 1) = A x + 1 − 2 2x − 1 b. 2x 2 − 3x (x 2 + 1)(3x + 2) = B 3x + 2 − 1 x 2 + 1 5. Evaluate the integral. a.  3 (x + 1)(1 − 2x) dx b.  2x 2 − 3x (x 2 + 1)(3x + 2) dx 436 CHAPTER 7 Principles of Integral Evaluation Exercise Set 7.5 CAS 1–8 Write out the form of the partial fraction decomposition. (Do not find the numerical values of the coefficients.) 1. 3x − 1 (x − 3)(x + 4) 2. 4 (x − 1)(x − 2)(x − 3) 3. 2x − 3 x 3 − x 2 4.

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  Maths by Van Rensburg N5 Students Book

  TVET FIRST

  • MJJ van Rensburg(Author)
  • 2022(Publication Date)
  • Macmillan

    (Publisher)

  Module 4: Integration techniques 107 TVET FIRST MODULE 4 Unit 4.5: Integration of algebraic fractions • First divide if the highest power in the numerator is greater than or equal to the highest power in the denominator: ¡ Step 1: Perform long division. ¡ Step 2: Use the solution from long division to write down the integral. ¡ Step 3: Apply the rules you have learned to determine the integral. Unit 4.6: Partial fractions • When we add fractions, they are reduced to a single fraction. a _ x + b _ y = a _ x × y _ y + b _ y × x _ x = ay + bx _ xy • We use the technique of partial fractions to break up a fraction into two or more fractions in order to integrate. • If the highest power of x in the numerator is greater than or equal to the highest power in the denominator, we must first divide. • Two distinct factors in the denominator ¡ When there are two distinct factors in the denominator, we need to add fractions where each has one of the factors as a denominator, such as in f (x) ____________ (ax ± b)(cx ± d) = A _ ax ± b + B _ cx ± d . • Integration Using Partial Fractions ¡ Integrate ∫ 3x + 2 _ 2 x 2 − x − 1 dx by breaking it up by means of partial fractions into two fractions that can be integrated. • Fractions with two or three recurring factors We can break up ∫ f (x) _ (ax ± b) n dx fractions as follows: ¡ 1 _ (x − 1) 2 = A _ x − 1 + B _ (x − 1) 2 ¡ 4 _ (2x + 1) 3 = A _ 2x + 1 + B _ (2x + 1) 2 + C _ (2x + 1) 3 ¡ 1 _ (ax + b) n = A _ ax + b + B _ (ax + b) 2 + C _ (ax + b) 3 + … + N _ (ax + b) n Step 1: Resolve into partial fractions: ¡ Equate the powers to determine numerators A and B. ¡ Use A and B to determine the constant. Step 2: Integrate using the partial fractions.

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  The Calculus Lifesaver

  All the Tools You Need to Excel at Calculus

  • Adrian Banner(Author)
  • 2009(Publication Date)
  • Princeton University Press

    (Publisher)

  18.3 Partial Fractions Let’s focus our attention on how to integrate a rational function. So we want to find an integral like Z p ( x ) q ( x ) dx, where p and q are polynomials. This covers a whole slew of integrals, for example, Z x 2 + 9 x 4 -1 dx, Z x x 3 + 1 dx, or Z 1 x 3 -2 x 2 + 3 x -7 dx. 398 • Techniques of Integration, Part One These seem a little complicated. Here are some simpler ones: Z 1 x -3 dx, Z 1 ( x + 5) 2 dx, Z 1 x 2 + 9 dx, and Z 3 x x 2 + 9 dx. The last four integrands are all rational functions, but they are a lot simpler. Try to work out all of these integrals using substitution. (Hint: some substi-tutions which work are t = x -3, t = x + 5, t = x/ 3, and t = x 2 + 9 for the four integrals, respectively.) The first two of these integrals have denomina-tors which are powers of linear functions, whereas the last two have quadratic denominators which cannot be factored. So, here’s the idea: first we’ll see how to take a general rational function and do some algebra to bust it up into a sum of simpler rational functions; then we’ll see how to integrate the simpler types of rational functions. The simpler functions I’m talking about are all like the four above: they either look like a constant over a linear power, or they look like a linear function over a quadratic. We’ll look at the algebra first, then the calculus. Finally, we’ll give a summary and look at a big example. 18.3.1 The algebra of partial fractions Our goal is to break up a rational function into simpler pieces. The first step in this process is to make sure that the numerator of the function has degree less than the denominator. If not, we’ll have to start off with a long division. So in the examples Z x + 2 x 2 -1 dx and Z 5 x 2 + x -3 x 2 -1 dx, the first is fine, since the degree of the top (1) is less than the degree of the bottom (2). The second example isn’t so great, because the degrees of the top and bottom are equal (to 2).

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  Calculus Early Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  54. Writing Suppose that P(x) is a cubic polynomial. State the general form of the partial fraction decomposition for f(x) = P(x) (x + 5) 4 and state the implications of this decomposition for evaluat- ing the integral ∫ f(x) dx. 55. Writing Consider the functions f(x) = 1 x 2 − 4 and g(x) = x x 2 − 4 Each of the integrals ∫ f(x) dx and ∫ g(x) dx can be eval- uated using partial fractions and using at least one other integration technique. Demonstrate two different techniques for evaluating each of these integrals, and then discuss the considerations that would determine which technique you would use. ✓Quick Check Answers 7.5 1. A 1 (ax + b) + A 2 (ax + b) 2 + ⋯+ A m (ax + b) m 2. a. A proper rational function is a rational func- tion in which the degree of the numerator is less than the degree of the denominator. b. The degree of the numerator must be less than the degree of the denominator. c. Divide the denominator into the numerator, which results in the sumof a polynomial and a proper rational function. 3. a. A 1 ax + b + A 2 (ax + b) 2 + ⋯ + A m (ax + b) m b. A 1 x + B 1 ax 2 + bx + c + A 2 x + B 2 (ax 2 + bx + c) 2 + ⋯ + A m x + B m (ax 2 + bx + c) m 4. a. A = 1 b. B = 1 5. a. ∫ 3 (x + 1)(1 − 2x) dx = ln | | | | x + 1 1 − 2x | | | | + C b. ∫ 2x 2 − 3x (x 2 + 1)(3x + 2) dx = 2 3 ln |3x + 2| − tan −1 x + C 484 Chapter 7 / Principles of Integral Evaluation 7.6 USING COMPUTER ALGEBRA SYSTEMS AND TABLES OF INTEGRALS In this section we will discuss how to integrate using tables, and we will see some special substitutions to try when an integral doesn’t match any of the forms in an integral table. In particular, we will discuss a method for integrating rational functions of sin x and cos x. We will also address some of the issues that relate to using computer algebra systems for integration. Readers who are not using computer algebra systems can skip that material. Integral Tables Tables of integrals are useful for eliminating tedious hand computation.

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  Calculus: Early Transcendentals, Metric Edition

  • James Stewart, Daniel K. Clegg, Saleem Watson, , James Stewart, James Stewart, Daniel K. Clegg, Saleem Watson(Authors)
  • 2020(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. SECTION 7.4 Integration of Rational Functions by Partial Fractions 515 7.4 Exercises 1–6 Write out the form of the partial fraction decomposition of the function (as in Example 7). Do not determine the numerical values of the coefficients. 1. (a) 1 s x 2 3ds x 1 5d (b) 2x 1 5 s x 2 2d 2 s x 2 1 2d 2. (a) x 2 6 x 2 1 x 2 6 (b) 1 x 2 1 x 4 3. (a) x 2 1 4 x 3 2 3x 2 1 2x (b) x 3 1 x x s2x 2 1d 2 s x 2 1 3d 2 4. (a) 5 x 4 2 1 (b) x 4 1 x 1 1 s x 3 2 1ds x 2 2 1d 5. (a) x 5 1 1 s x 2 2 xds x 4 1 2x 2 1 1d (b) x 2 x 2 1 x 2 6 6. (a) x 6 x 2 2 4 (b) x 4 s x 2 2 x 1 1ds x 2 1 2d 2 7–40 Evaluate the integral. 7. y 5 s x 2 1ds x 1 4d dx 8. y x 2 12 x 2 2 4x dx 9. y 5x 1 1 s2x 1 1ds x 2 1d dx 10. y y s y 1 4ds2y 2 1d dy 11. y 1 0 2 2x 2 1 3x 1 1 dx 12. y 1 0 x 2 4 x 2 2 5x 1 6 dx 13. y 1 x s x 2 ad dx 14. y 1 s x 1 ads x 1 bd dx 15. y x 2 x 2 1 dx 16. y 3t 2 2 t 1 1 dt 17. y 2 1 4y 2 2 7y 2 12 ys y 1 2ds y 2 3d dy 18. y 2 1 3x 2 1 6x 1 2 x 2 1 3x 1 2 dx 19. y 1 0 x 2 1 x 1 1 s x 1 1d 2 sx 1 2d dx 20. y 3 2 x s3 2 5xd s3x 2 1ds x 2 1d 2 dx 21. y dt st 2 2 1d 2 22. y 3x 2 1 12x 2 20 x 4 2 8x 2 1 16 dx 23. y 10 s x 2 1ds x 2 1 9d dx 24. y 3x 2 2 x 1 8 x 3 1 4x dx 25. y 0 21 x 3 2 4x 1 1 x 2 2 3x 1 2 dx 26. y 2 1 x 3 1 4x 2 1 x 2 1 x 3 1 x 2 dx 27. y 4x x 3 1 x 2 1 x 1 1 dx 28. y x 2 1 x 1 1 s x 2 1 1d 2 dx 29. y x 3 1 4x 1 3 x 4 1 5x 2 1 4 dx 30. y x 3 1 6x 2 2 x 4 1 6x 2 dx 31. y x 1 4 x 2 1 2x 1 5 dx 32. y 1 0 x x 2 1 4x 1 13 dx 33. y 1 x 3 2 1 dx 34. y x 3 2 2x 2 1 2x 2 5 x 4 1 4x 2 1 3 dx 35.

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  Single Variable Calculus, Metric Edition

  • James Stewart, Daniel K. Clegg, Saleem Watson, , James Stewart, James Stewart, Daniel K. Clegg, Saleem Watson(Authors)
  • 2020(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. SECTION 7.4 Integration of Rational Functions by Partial Fractions 553 7.4 Exercises 1–6 Write out the form of the partial fraction decomposition of the function (as in Example 7). Do not determine the numerical values of the coefficients. 1. (a) 1 s x 2 3ds x 1 5d (b) 2x 1 5 s x 2 2d 2 s x 2 1 2d 2. (a) x 2 6 x 2 1 x 2 6 (b) 1 x 2 1 x 4 3. (a) x 2 1 4 x 3 2 3x 2 1 2x (b) x 3 1 x x s2x 2 1d 2 s x 2 1 3d 2 4. (a) 5 x 4 2 1 (b) x 4 1 x 1 1 s x 3 2 1ds x 2 2 1d 5. (a) x 5 1 1 s x 2 2 xds x 4 1 2x 2 1 1d (b) x 2 x 2 1 x 2 6 6. (a) x 6 x 2 2 4 (b) x 4 s x 2 2 x 1 1ds x 2 1 2d 2 7–40 Evaluate the integral. 7. y 5 s x 2 1ds x 1 4d dx 8. y x 2 12 x 2 2 4x dx 9. y 5x 1 1 s2x 1 1ds x 2 1d dx 10. y y s y 1 4ds2y 2 1d dy 11. y 1 0 2 2x 2 1 3x 1 1 dx 12. y 1 0 x 2 4 x 2 2 5x 1 6 dx 13. y 1 x s x 2 ad dx 14. y 1 s x 1 ads x 1 bd dx 15. y x 2 x 2 1 dx 16. y 3t 2 2 t 1 1 dt 17. y 2 1 4y 2 2 7y 2 12 ys y 1 2ds y 2 3d dy 18. y 2 1 3x 2 1 6x 1 2 x 2 1 3x 1 2 dx 19. y 1 0 x 2 1 x 1 1 s x 1 1d 2 sx 1 2d dx 20. y 3 2 x s3 2 5xd s3x 2 1ds x 2 1d 2 dx 21. y dt st 2 2 1d 2 22. y 3x 2 1 12x 2 20 x 4 2 8x 2 1 16 dx 23. y 10 s x 2 1ds x 2 1 9d dx 24. y 3x 2 2 x 1 8 x 3 1 4x dx 25. y 0 21 x 3 2 4x 1 1 x 2 2 3x 1 2 dx 26. y 2 1 x 3 1 4x 2 1 x 2 1 x 3 1 x 2 dx 27. y 4x x 3 1 x 2 1 x 1 1 dx 28. y x 2 1 x 1 1 s x 2 1 1d 2 dx 29. y x 3 1 4x 1 3 x 4 1 5x 2 1 4 dx 30. y x 3 1 6x 2 2 x 4 1 6x 2 dx 31. y x 1 4 x 2 1 2x 1 5 dx 32. y 1 0 x x 2 1 4x 1 13 dx 33. y 1 x 3 2 1 dx 34. y x 3 2 2x 2 1 2x 2 5 x 4 1 4x 2 1 3 dx 35.

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