Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 93. A major complaint of professors teaching calculus is that students don't have the appropriate background to work through the calculus course successfully. This text is targeted directly at this underprepared audience. This is a single-variable (2-semester) calculus text that incorporates a conceptual re-introduction to key precalculus ideas throughout the exposition as appropriate. This is the ideal resource for those schools dealing with poorly prepared students or for schools introducing a slower paced, integrated precalculus/calculus course | Integration Using Partial Fractions 901 EXAMPLE Find f x2 2x2-i-1 Jx SOLUTION The integrand can be broken into a sum as follows 2- 1 _ A B Cx D x2 x2 1 x x2 x2 1 A repeated factor contributes multiple fractions as We can solve for A B C and D. One method is to clear denominators and evaluate both sides at four different x-values. This leaves us with four equations and four unknowns. We can show that 2x 1 _ 2 1 2x 1 x2 x2 1 x x2 x2 1 Jx 2 f 1 Jx f Jx f 2 1 Jx 2 x2 1 x x2 x2 1 1 2x 1 2 ln x -------- I --------- Jx I ------ Jx 11 1 x2 1 x2 1 _ . 1 X 2. 2 ln x ln x 1 arctan x C x General Procedure for Partial Fraction Decomposition of Qx If QU is improper do long division to express it as the sum of a polynomial and a proper rational function. Factor Q x into the product of linear and irreducible quadratic factors. x a is a factor of Q x if 2 a 0 Rewrite the proper rational function as a sum of simpler rational functions as follows. For every nonrepeated linear factor x a of Q there is a term For a repeated linear factor x a of Q there are n partial fractions. All numerators are constant denominators consist of x a raised to successively higher powers. . for x - a 3 a factor of Q there are fractions B 2 C 3. a x a 2 x a 3 For a nonrepeated irreducible quadratic factor of Q x2 fex c there s a term Ax B x2 fex c . If the quadratic factor of Q is repeated there are multiple partial fractions each with a linear expression in the numerator. . x2 4 2 as a factor of Q would result in the partial fractions A2 B C9x .D9. 2 4 x2 4 2 Finding numerical values for the constants is an algebra problem. Integrating the partial fractions coming from linear factors of Q is straightforward. Some quadratics and repeated quadratics are straightforward while others require both completing the square and trigonometric substitution. 3 The denominators are x and x2 not x and x. 902 CHAPTER 29 Computing Integrals PROBLEMS FOR SECTION In Problems 1 through 6 write .
