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## TECHNIQUES OF INTEGRATION

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**7**TECHNIQUES OF INTEGRATION**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.**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.**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:**INTEGRATION BY PARTIAL FRACTIONS**If we now reverse the procedure, we see how to integrate the function on the right side of this equation:**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.**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.**DEGREE OF P**Recall that, if where an≠ 0, then the degree of P is n and we write deg(P) = n.**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).**PARTIAL FRACTIONS**Equation 1 The division statement is where S and R are also polynomials.**PARTIAL FRACTIONS**As the following example illustrates, sometimes, this preliminary step is all that is required.**PARTIAL FRACTIONS**Example 1 Find • The degree of the numerator is greater than that of the denominator. • So, we first perform the long division.**PARTIAL FRACTIONS**Example 1 • This enables us to write:**PARTIAL FRACTIONS**The next step is to factor the denominator Q(x) as far as possible.**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).**FACTORISATION OF Q(x)**For instance, if Q(x) = x4 – 16, we could factor it as:**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:**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.**CASE 1**The denominator Q(x) is a product of distinct linear factors.**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.**CASE 1**Equation 2 In this case, the partial fraction theorem states that there exist constants A1, A2, . . . , Ak such that:**CASE 1**These constants can be determined as in the following example.**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.**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.**PARTIAL FRACTIONS**E. g. 2—Equation 3 So, the partial fraction decomposition of the integrand (Equation 2) has the form**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)**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**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.**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**PARTIAL FRACTIONS**Example 2 Solving, we get: • A = ½ • B = 1/5 • C = –1/10**PARTIAL FRACTIONS**Example 2 Hence,**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.**NOTE**We can use an alternative method to find the coefficients A, B, and Cin Example 2.**NOTE**Equation 4 is an identity. It is true for every value of x. • Let’s choose values of x that simplify the equation.**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.**NOTE**Likewise, x = ½ gives 5B/4 = 1/4 and x = –2 gives 10C = –1. • Hence, B = 1/5 and C = –1/10.**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?**NOTE**In fact, Equation 4 is true for all values of x, even x = 0, ½, and –2 . • See Exercise 69 for the reason.**PARTIAL FRACTIONS**Example 3 Find , where a≠ 0. • The method of partial fractions gives: • Therefore,**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).**PARTIAL FRACTIONS**Example 3 Therefore,**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.**CASE 2**Q(x) is a product of linear factors, some of which are repeated.**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).**CASE 2**Equation 7 Then, instead of the single term A1/(a1x +b1) in Equation 2, we would use:**CASE 2**By way of illustration, we could write: • However, we prefer to work out in detail a simpler example, as follows.**PARTIAL FRACTIONS**Example 4 Find • The first step is to divide. • The result of long division is:**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:**PARTIAL FRACTIONS**Example 4 The linear factor x – 1 occurs twice. So, the partial fraction decomposition is:**PARTIAL FRACTIONS**E. g. 4—Equation 8 Multiplying by the least common denominator, (x – 1)2 (x + 1), we get: