MTH 252 Integral Calculus Chapter 8 Principles of Integral Evaluation Section 8 5 Integrating Rational Functions by Partial Fractions Copyright 2006 by Ron Wallace all rights reserved Review Addition Subtraction of Fractions a c ad bc b d bd Note The equation works both ways Problem Find two fractions whose sum difference is equal to a third given fraction The product of the denominators of the two fractions will be the denominator of the given fraction Example ad bc a c bd b d Find two fractions whose sum or difference is 11 35 Denominator 35 Possibilities for the other two denominators are 1 35 and 5 7 11 A B 35 5 7 11 7 A 5B Many solutions including 3 2 and 1 4 5 Rational Functions Any function of the form P x f x Q x where P x and Q x are polynomials Fundamental Theorem of Algebra Every polynomial equation of degree n with complex coefficients has n roots in the complex numbers Each real root r gives a factor x r Each complex root i has a companion root i These give a factor ax2 bx c Hence every polynomial can be written as a product of linear quadratic factors Some Easy Integrals of Rational Functions Linear Denominator 5 dx 5ln x 3 c x 3 5 dx 2x 3 Let u 2x 3 du 2dx 5 1 5 du ln 2 x 3 c 2 u 2 4x 12 dx 4 dx 4 x 12 ln x 3 c x 3 x 3 17 4 x 5 4 dx 4 x 17 ln x 3 c x 3 dx x 3 5 dx 7 2 x 3 Let u 2x 3 1 52 7 du 52 u 7 du u du 2dx 5 12 u 6 c 5 12 2 x 3 6 c Partial Fractions P x dx Q x where Q x is a product of linear factors deg P x deg Q x 2 x 5 A B 3x 1 x 4 3 x 1 x 4 2 x 5 A x 4 B 3 x 1 A 3B x 4 A B 2 A 3B 5 4 A B Solve this system for A B A 1 B 1 2 x 5 1 1 1 dx dx ln 3x 1 ln x 4 c 3x 1 x 4 3 x 1 x 4 3 This method can be extended to any number of distinct linear factors Partial Fractions Repeated Linear Factors P x dx Q x 4x 1 2 x 3 2 where Q x is a product of linear factors deg P x deg Q x A BB 2 x 3 2 x 3 2 4 x 1 A 2 x 3 B 2 Ax 3 A B 4 2 A 1 3 A B Solve this system for A B A 2 B 7 4x 1 2 7 7 dx dx ln 2 x 3 c 2 2 2 x 3 2 x 3 2 x 3 2 2 x 3 This method can be extended to any power of the denominator and can be combined with the previous method More Easy Integrals of Rational Functions Quadratic Denominator 3 1 dx 3 dx 2 2 x 4 x 7 x 2 3 1 3 2 du u 3 Let u x 2 du dx Finish using trig substitutions 3 2x 4 17 3 x 11 dx 2 dx dx 2 2 2 x 4 x 7 x 4 x 7 x 4 x 7 3 6 11 2 3 x 11 2 x 4 d 2 x 4 x 7 dx Just like the one above 3 2 ln x 2 4 x 7 More Easy Integrals of Rational Functions Quadratic Denominator 3 2x 4 17 3x 11 dx dx dx 2 5 2 5 2 5 2 x 4 x 7 x 4 x 7 x 4 x 7 Complete the square trig substitution Let u x2 4x 7 du 2x 4 dx 3 5 3 4 u du u c 2 8 3 2 x 4 x 7 4 c 8 Partial Fractions x 2 3 x 2 x 1 x 2 2 x 3 P x dx Q x where Q x is a product of a linear factor a quadratic factor deg P x deg Q x A BxB C 2 x 1 x 2 x 3 x 2 3x 2 A B x 2 2 A B C x 3 A C 1 A B 3 2 A B C 2 3 A C Solve this system for A B C A 1 B 0 C 1 x 2 3x 2 1 1 dx dx 2 dx 2 x 1 x 2 x 3 x 1 x 2 x 3 This method can be extended to any number of distinct linear quadratic factors Partial Fractions Repeated Quadratic Factors P x dx Q x where Q x is a product of quadratic factors deg P x deg Q x P x 3 x 2 x 2 2 Ax B Cx D 2 2 3x x 2 3 x x 2 2 and proceed as before All of these methods can be combined and extended to handle any rational function where you can factor the denominator into a product of linear and quadratic factors P x dx Q x where deg P x deg Q x Simplify using long division of polynomials Example 4 x 3 3x 5 2 x 1 dx 2 4 2 x dx xdx dx dx 2x 1 2 4 2x 1 2x x 1 2 x 1 4 x 3 0 x 2 3x 5 4 x3 2 x2 2 x 2 3x 2x 2 x 2 x 5 2 x 1 4