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TECHNIQUES OF INTEGRATION

Learn how to integrate rational functions by expressing them as a sum of simpler fractions using the technique of partial fractions.

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TECHNIQUES OF INTEGRATION

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  1. 7 TECHNIQUES OF INTEGRATION

  2. TECHNIQUES OF INTEGRATION 7.4Integration of Rational Functions by Partial Fractions • In this section, we will learn: • How to integrate rational functions • by reducing them to a sum of simpler fractions.

  3. PARTIAL FRACTIONS • We show how to integrate any rational function (a ratio of polynomials) by expressing it as a sum of simpler fractions, called partial fractions. • We already know how to integrate partial functions.

  4. INTEGRATION BY PARTIAL FRACTIONS • To illustrate the method, observe that, by taking the fractions 2/(x – 1) and 1/(x – 2) to a common denominator, we obtain:

  5. INTEGRATION BY PARTIAL FRACTIONS • If we now reverse the procedure, we see how to integrate the function on the right side of this equation:

  6. INTEGRATION BY PARTIAL FRACTIONS • To see how the method of partial fractions works in general, let’s consider a rational function • where P and Q are polynomials.

  7. PROPER FUNCTION • It’s possible to express f as a sum of simpler fractions if the degree of P is less than the degree of Q. • Such a rational function is called proper.

  8. DEGREE OF P • Recall that, if • where an≠ 0, then the degree of P is n and we write deg(P) = n.

  9. PARTIAL FRACTIONS • If f is improper, that is, deg(P) ≥ deg(Q), then we must take the preliminary step of dividing Q into P (by long division). • This is done until a remainder R(x) is obtained such that deg(R) < deg(Q).

  10. PARTIAL FRACTIONS Equation 1 • The division statement is • where S and R are also polynomials.

  11. PARTIAL FRACTIONS • As the following example illustrates, sometimes, this preliminary step is all that is required.

  12. PARTIAL FRACTIONS Example 1 • Find • The degree of the numerator is greater than that of the denominator. • So, we first perform the long division.

  13. PARTIAL FRACTIONS Example 1 • This enables us to write:

  14. PARTIAL FRACTIONS • The next step is to factor the denominator Q(x) as far as possible.

  15. FACTORISATION OF Q(x) • It can be shown that any polynomial Qcan be factored as a product of: • Linear factors (of the form ax +b) • Irreducible quadratic factors (of the form ax2 + bx +c, where b2 – 4ac < 0).

  16. FACTORISATION OF Q(x) • For instance, if Q(x) = x4 – 16, we could factor it as:

  17. FACTORISATION OF Q(x) • The third step is to express the proper rational function R(x)/Q(x) as a sum of partial fractionsof the form:

  18. FACTORISATION OF Q(x) • A theorem in algebra guarantees that it is always possible to do this. • We explain the details for the four cases that occur.

  19. CASE 1 • The denominator Q(x) is a product of distinct linear factors.

  20. CASE 1 • This means that we can write • Q(x) = (a1x + b1) (a2x + b2)…(akx + bk) • where no factor is repeated (and no factor is a constant multiple of another.

  21. CASE 1 Equation 2 • In this case, the partial fraction theorem states that there exist constants A1, A2, . . . , Ak such that:

  22. CASE 1 • These constants can be determined as in the following example.

  23. PARTIAL FRACTIONS Example 2 • Evaluate • The degree of the numerator is less than the degree of the denominator. • So, we don’t need to divide.

  24. PARTIAL FRACTIONS Example 2 • We factor the denominator as: 2x3 + 3x2 – 2x = x(2x2 + 3x – 2) = x(2x – 1)(x + 2) • It has three distinct linear factors.

  25. PARTIAL FRACTIONS E. g. 2—Equation 3 • So, the partial fraction decomposition of the integrand (Equation 2) has the form

  26. PARTIAL FRACTIONS E. g. 2—Equation 4 • To determine the values of A, B, and C, we multiply both sides of the equation by the product of the denominators, x(2x – 1)(x + 2), obtaining: • x2 + 2x + 1 = A(2x – 1)(x + 2) + Bx(x + 2) • + Cx(2x – 1)

  27. PARTIAL FRACTIONS E. g. 2—Equation 5 • Expanding the right side of Equation 4 and writing it in the standard form for polynomials, we get: • x2 + 2x + 1 = (2A + B + 2C)x2 • + (3A + 2B – C) – 2A

  28. PARTIAL FRACTIONS Example 2 • The polynomials in Equation 5 are identical. • So, their coefficients must be equal. • The coefficient of x2 on the right side, 2A +B + 2C, must equal that of x2 on the left side—namely, 1. • Likewise, the coefficients of x are equal and the constant terms are equal.

  29. PARTIAL FRACTIONS Example 2 • This gives the following system of equations for A, B, and C: • 2A + B + 2C = 1 • 3A + 2B – C = 2 • –2A = –1

  30. PARTIAL FRACTIONS Example 2 • Solving, we get: • A = ½ • B = 1/5 • C = –1/10

  31. PARTIAL FRACTIONS Example 2 • Hence,

  32. PARTIAL FRACTIONS Example 2 • In integrating the middle term, we have made the mental substitution u = 2x – 1, which gives du = 2 dx and dx =du/2.

  33. NOTE • We can use an alternative method to find the coefficients A, B, and Cin Example 2.

  34. NOTE • Equation 4 is an identity. • It is true for every value of x. • Let’s choose values of x that simplify the equation.

  35. NOTE • If we put x = 0 in Equation 4, the second and third terms on the right side vanish, and the equation becomes –2A = –1. • Hence, A = ½.

  36. NOTE • Likewise, x = ½ gives 5B/4 = 1/4 and x = –2 gives 10C = –1. • Hence, B = 1/5 and C = –1/10.

  37. NOTE • You may object that Equation 3 is not valid for x = 0, ½, or –2. • So, why should Equation 4 be valid for those values?

  38. NOTE • In fact, Equation 4 is true for all values of x, even x = 0, ½, and –2 . • See Exercise 69 for the reason.

  39. PARTIAL FRACTIONS Example 3 • Find , where a≠ 0. • The method of partial fractions gives: • Therefore,

  40. PARTIAL FRACTIONS Example 3 • We use the method of the preceding note. • We put x = a in the equation and get A(2a) = 1. So, A = 1/(2a). • If we put x = –a, we get B(–2a) = 1. So, B =–1/(2a).

  41. PARTIAL FRACTIONS Example 3 • Therefore,

  42. PARTIAL FRACTIONS E. g. 3—Formula 6 • Since ln x –ln y =ln(x/y), we can write the integral as: • See Exercises 55–56 for ways of using Formula 6.

  43. CASE 2 • Q(x) is a product of linear factors, some of which are repeated.

  44. CASE 2 • Suppose the first linear factor (a1x +b1) is repeated r times. • That is, (a1x +b1)r occurs in the factorization of Q(x).

  45. CASE 2 Equation 7 • Then, instead of the single term A1/(a1x +b1) in Equation 2, we would use:

  46. CASE 2 • By way of illustration, we could write: • However, we prefer to work out in detail a simpler example, as follows.

  47. PARTIAL FRACTIONS Example 4 • Find • The first step is to divide. • The result of long division is:

  48. PARTIAL FRACTIONS Example 4 • The second step is to factor the denominator Q(x) = x3 – x2 – x + 1. • Since Q(1) = 0, we know that x – 1 is a factor, and we obtain:

  49. PARTIAL FRACTIONS Example 4 • The linear factor x – 1 occurs twice. • So, the partial fraction decomposition is:

  50. PARTIAL FRACTIONS E. g. 4—Equation 8 • Multiplying by the least common denominator, (x – 1)2 (x + 1), we get:

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