If deg(num) deg(denom), then we should use long-division of polynomials to rewrite: p(x) = s(x) + r(x) q(x), q(x)
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1 Mth 50 The method of prtil frction decomposition (PFD is used to integrte some rtionl functions of the form p(x, where p/q is in lowest terms nd deg(num < deg(denom. q(x If deg(num deg(denom, then we should use long-division of polynomils to rewrite: p(x q(x = s(x + r(x q(x, where s(x is the quotient, r(x is the reminder, nd deg(r < deg(q. In this cse, we would hve where p(x q(x = s(x + r(x q(x, s(x is esy becuse s is polynomil nd we might pply PFD to r(x q(x. Before ttempting PFD, we should check to see if substitution is more vible option s it is typiclly esier to perform thn PFD. There re two cses to consider when performing PFD: If the denomintor q(x hs liner fctor (x c of multiplicity m, where c is rel number, then the sum will hve terms of the form where ech A i is constnt. A x c + A (x c + A (x c + + A m (x c m, If the denomintor q(x hs n irreducible qudrtic fctor (x +bx+c of multiplicity m, tht is, if b c < 0, then the sum will hve terms of the form B x + C x + bx + c + B x + C (x + bx + c + + B mx + C m (x + bx + c, m where ech B i nd C i is constnt. Another tool tht is often helpful when working integrls tht involve PFD is completing the squre: Every qudrtic function f(x = x + bx + c cn be written s f(x = (x h + k, where h = b nd k = f(h.
2 Mth 50 EXAMPLE #. Solve x x + x 8. Solution. ( Fctor the denomintor: Using the AC-Method, we hve x +x 8 = (x+(x. ( Form the PFD: We hve distinct liner fctors, so the integrnd cn be decomposed s x (x + (x = A x + + B x. A(x + B(x + = x. ( ( Find the constnts A nd B: (Method I Solve the system tht results from expnding the identity (*. We hve (A+Bx+( A+B = x, which mens tht A+B = nd A+B = 0. We could use substitution, elimintion, or perform elementry row opertions to solve this system: A = nd B =. Those who hve hd some experience with liner lgebr might pprecite this method. (Method Exploit the identity (* by plugging in vlues for x tht simplify the expression. If x =, then A(0 + B(6 = 6, so tht B =. If x =, then A( 6 + B(0 =, so tht A =. If this method works, it is reltively esy to crry-out. ( Solve the integrl: ( x x + x 8 = x + + = x x + + x = ln x + + ln x + C. REMARK. We hve invested lot of time/effort into decomposing the integrnd; however, the ctul work of integrtion is now reltively esy. EXAMPLE #. Solve x x x + 8. Solution. ( Fctor the denomintor: Using fctor by grouping, we hve x x x + 8 = x (x (x = (x + (x. ( Form the PFD: We hve some repeted liner fctors, so the integrnd cn be decomposed s (x + (x = A x + + B x + C (x. A(x + B(x + (x + C(x + =. (
3 Mth 50 ( Find the constnts A, B, nd C: Let s exploit the identity (* by plugging in vlues for x tht simplify the expression. If x =, then C( =, so tht C = /. If x =, then A( =, so tht A = /6. To find B, note tht by expnding some of (*, we get Ax + Bx = 0x, so B = A = /6. Sometimes, the best wy to find the constnts is combintion of the two methods! ( Solve the integrl: ( /6 x x x + 8 = x + + /6 x + / (x = 6 x + 6 = 6 ln x + 6 ln x + x + (x ( (x + C = 6 ln x + 6 ln x (x + C. In the third line bove, I used substitution with p = x, nd then I returned to the originl vribles. A Useful Arctngent Theorem. Recll tht, where 0? We hve tht + x + x = ( + x = + x = tn (x + C. Well, then, wht is + ( x. Let p = x, so tht dp =, or = dp. + ( x = + p dp = tn (p + C = ( x tn + C. Therefore, + x = ( x tn + C.
4 Mth 50 EXAMPLE #. Solve x x. Solution. long-division: Since deg(num deg(denom, before we cn use PFD, we will need to perform x x = x + x x. x We now perform PFD on x. ( Fctor the denomintor: Using the difference of perfect cubes formul, we hve x = (x (x + x +. ( Form the PFD: We hve liner fctor nd n irreducible qudrtic fctor, giving the following decomposition: x (x (x + x + = A x + Bx + C x + x +. A(x + x + + (Bx + C(x = x. ( ( Find the constnts A, B, nd C: If x =, then A( =, so tht A =. To find B, note tht Ax + Bx = 0x, so B = A =. To find C, note tht A C = 0, so C = A =. ( Solve the integrl: x ( x = x + x x = ( x + x + x +. x + x + The first two pieces of this integrl re esy: x = x + C ( If we cn solve x = ln x + C. ( x, we re done! Completing the squre on the denomintor gives x + x + x + x + = ( x + +.
5 Mth 50 Let s use the substitution p = x +. dp = nd x = p. This gives x x + x + = ( p p + dp = ( p p + = = ( dp ( dp p + tn ( p p ( dp p + q dq, where we used the Useful Arctngent Theorem with = / nd the substitution q = p + ( /, so tht dq = p dp, or p dp = dq. Further simplifiction nd returning to the originl vribles gives x x + x + = ( tn (x / ( ln p + / + C = tn ( (x ln x + x + + C. ( Putting Equtions (, (, nd ( together gives us the finl result: x x = x + ln x + ( tn (x ln x + x + + C. WHEW! This reltively simple looking integrnd hs n ntiderivtive tht is just crzy. NOTE: We didn t work problem with repeted irreducible qudrtic fctors, but the ide is very similr to wht we did with the repeted liner fctors. For exmple, we hve the following PFD form: f(x = x x + x + 5 x (x + x + 7 = A x + B x + C x + Dx + E x + x F x + G (x + x + 7, which would result in system of SEVEN EQUATIONS IN SEVEN UNKNOWNS, just to find the constnts! The first three pieces would be esy to integrte, but the rest of it... mybe tht s job more suited for Mple. Indeed, Mple hs built-in procedure for PFD, convert(f,prfrc, where f is the integrnd. Try it for the bove function f(x nd then use Mple to integrte! 5
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