Partial Fractions

3 Key excerpts on "Partial Fractions"

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  eBook - PDF

  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 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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  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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