4<‡Mze%$#"!  ijklmnopqrstuvwxyz{|}~€‚ƒ„…†‡†<;:9876ZLī okBhcPartial FractionskBhc(!Introduction to Partial FractionskBhc,%Given a rational function of the formcarFTkBhcpGrapkBhcxkBhcqGrapkBhcxkBhcwhere the degree of kBhcpGrapkBhcxkBhc# is less than the degree of kBhcqGrapkBhcxkBhc’, the method of partial fractions seeks to break this rational function down into the sum of simpler rational functions. In particular, we are going to try to write the original function as a sum of rational functions where the degrees of the polynomials kBhc)"involved are as small as possible.kBhc@9The steps in the partial fractions method are as follows:kBhc“ŒMake sure that the degree of the numerator is less than the degree of the denominator. (Below we will see what to do when that is not true.) kBhc3,Factor the denominator as fully as possible. kBhc’Write the original rational function as a sum of fractions. Each of the terms in the sum gets one of the original factors as its denominator. The numerator of each new fraction is an arbitrary polynomial of degree one less than the degree of the denominatkBhc or. kBhcXQUse algebra to solve for the coefficients in the numerators of the new fractions. kBhcVOHere is a step by step tour through the method applied to the rational function carFWkBhc2 x + 4kBhcxopxEkBhc2kBhc - 4 x - 5kBhctmNote that the degree of the numerator is one less than the degree of the denominator, as required in 1 above.kBhc92Next, we factor the polynomial in the denominator.carFWkBhc2 x + 4kBhcxopxEkBhc2kBhc - 4 x - 5kBhc  = carFVkBhc2 x + 4GrapkBhc x - 5GrapkBhc x + 1kBhcõīWe then attempt to write the fraction as a sum of fractions, with each fraction getting one of the terms in the factored form. The numerator of each fraction is a polynomial with degree one less than that of polynomial in the denominator.carFVkBhc2 x + 4GrapkBhc x - 5GrapkBhc x + 1kBhc  = carF(kBhcAkBhc x - 5kBhc  + carF(kBhcBkBhc x + 1kBhc‡€To solve for the unknown coefficients on the right, we multiply both sides by the denominator of the original rational function.kBhc 4 + 2 x = GrapkBhc -5 + xkBhc GrapkBhc 1 + xkBhc Grap‡carF5kBhcAGrapkBhc -5 + xkBhc  + carF4kBhcBGrapkBhc 1 + xkBhc;4We get some nice cancelation on the right, giving uskBhc 4 + 2 x = A GrapkBhc 1 + xkBhc  + B GrapkBhc x - 5 kBhc@9There are two ways to solve for the unknown coefficients kBhcAkBhc  and kBhcBkBhcF?. The simplest way is to make some judicious substitutions for kBhcxkBhc . If we set kBhc x = -1kBhc , feR (1)kBhc becomeskBhc 2 = - 6 BkBhc orkBhc B = - carF$kBhc1kBhc3kBhc If we set kBhc x = 5kBhc , feR (1)kBhc becomeskBhc14 = 6 AkBhc orkBhc A = carF$kBhc7kBhc3kBhc˜‘An alternative technique is to expand out the expressions on the right and regroup terms to make the right hand side look like a polynomial in x. kBhc 4 + 2 x = GrapkBhc A + BkBhc x + A - 5 B!kBhckdComparing like terms on either side of the equation gives us a set of two equations in two unknowns."kBhc 2 = A + B#kBhc 4 = A - 5 B$kBhcg`I am assuming that you all know how to solve systems of equations, so I will skip to the answer.%kBhc A = carF$kBhc7kBhc3&kBhc B = carF%kBhc -1kBhc3'kBhc)"By either method, we now can write(carFVkBhc2 x + 4GrapkBhc x - 5GrapkBhc x + 1kBhc  = carF*kBhc 7/3kBhc x - 5kBhc  - carF*kBhc 1/3kBhc x + 1)kBhcŌĶIf the problem is to integrate the original rational function, it is easy to see that it is many times easier to integrate the two simpler rational functions that arise from the partial fraction expansion.*etnIzcarFVkBhc2 x + 4GrapkBhc x - 5GrapkBhc x + 1kBhcxkBhc  = etnINcarF*kBhc 7/3kBhc x - 5kBhcxkBhc  - etnINcarF*kBhc 1/3kBhc x + 1kBhcx+kBhcThe substitution,kBhc u = x - 5-kBhc$turns the first integral into.etnIdcarF@carF$kBhc7kBhc3kBhcukBhcukBhc  = carF$kBhc7kBhc3kBhc ln u + C/kBhc Hence0etnINcarF*kBhc 7/3kBhc x - 5kBhcxkBhc  = carF$kBhc7kBhc3kBhc  lnGrapkBhc x - 5kBhc  + C1kBhc!Likewise, the substitution2kBhc u = x + 13kBhc&turns the second integral into 4kBhc - etnIdcarF@carF$kBhc1kBhc3kBhcukBhcukBhc  = - carF$kBhc1kBhc3kBhc ln u + C = - carF$kBhc1kBhc3kBhc  lnGrapkBhc x + 15kBhc&Putting this all together gives6 etnI†carFbkBhc2 x + 4GrapkBhc x - 5kBhc GrapkBhc x + 1kBhcxkBhc  = carF$kBhc7kBhc3kBhc  lnGrapkBhc x - 5kBhc  - carF$kBhc1kBhc3kBhc  ln GrapkBhc x + 1kBhc  + ChkBhcAnother examplefkBhc#Suppose we have to integratedcarFQkBhc3 x + 2kBhcxopxEkBhc3kBhc  - 1ckBhcHAThe first step is to factor the denominator as fully as possible.bcarFQkBhc3 x + 2kBhcxopxEkBhc3kBhc  - 1kBhc  = carF}kBhc3 x + 2GrapkBhc x - 1Grap?kBhcxopxEkBhc2kBhc + x + 1akBhcxWe then write the fraction as a sum of simpler fractions whose denominators are the factors of the original denominator.`carF}kBhc3 x + 2GrapkBhc x - 1Grap?kBhcxopxEkBhc2kBhc + x + 1kBhc  = carF(kBhcAkBhc x - 1kBhc  + carFUkBhcB x + CkBhcxopxEkBhc2kBhc + x + 1_kBhcRKWe next multiply both sides by the denominator of the fraction on the left.^kBhc 3 x + 2 = A Grap?kBhcxopxEkBhc2kBhc + x + 1kBhc  + GrapkBhcB x + CGrapkBhc x - 1]kBhc~wThere are two methods for how to procede next. The first method is to substitute several different possible values for kBhcxkBhc in an effort to get feR ^(2)kBhc0) to reduce to a set of simpler equations.\ kBhc-&The first place to look for values of kBhcxkBhc“­ to substitute is to pick values that cause one or more of the factors on the right to vanish. In this example, the only way to make something vanish on the right is to set kBhcxkBhc kBhc=kBhc 1. Substituting kBhcxkBhc kBhc=kBhc 1 gives[kBhc5 = 3 AZkBhcA = 5/3Y kBhc³¬Once we have used up all of the substitutions that make terms vanish, we next try substitutions that are likely to yeild simple results. A popular choice for this stage is kBhcxkBhc kBhc=kBhc 0. Substituting kBhcxkBhc kBhc=kBhc 0 givesXkBhc 2 = A - CWkBhcC = -1/3VkBhc/(Finally, another simple substitution is kBhc x = -1kBhc:UkBhc-1 = A - 2(C - B)TkBhcB = -5/3SkBhc1*A second method for dealing with equation feR ^(2)kBhcž— is to expand the terms on the right hand side and then compare terms with like powers on either side of the equation. This leads to a set of equationsRtSqeTkBhc 0 = A + BkBhc 3 = A - B + CkBhc 2 = A - CQkBhc‡€We then have to solve that system of equations by some means. I will not show the details, but the result is the same as before:PtSqe»kBhc A = carF$kBhc5kBhc3kBhc B = carF%kBhc -5kBhc3kBhc C = carF%kBhc -1kBhc3OkBhc]VNow that we have determined the coefficients, we can move on to solving the integrals.NetnIucarFQkBhc3 x + 2kBhcxopxEkBhc3kBhc  - 1kBhcxkBhc  = etnIhcarFDcarF$kBhc5kBhc3kBhc x - 1kBhcxkBhc  + etnIŌcarF°kBhc - carF$kBhc5kBhc3kBhc  x - carF$kBhc1kBhc3kBhcxopxEkBhc2kBhc + x + 1kBhcxMkBhc`YBoth integrals can be handled via a simple substitution. I will leave the details to you.LkBhc=6What to do if the degree of the numerator is too largeKkBhctmThe next example shows what to do when the numerator of the rational function has a degree which is too high.JetnI–carFrkBhcxopxEkBhc3kBhc  + xkBhcxopxEkBhc2kBhc  + 2kBhcxIkBhcÕĪBefore we apply partial fractions, we have to apply the polynomial equivalent of long division, which works very much like long division for numbers. Consider this example. What can you do with the fractionHcarF%kBhc 21kBhc5GkBhcohto ŌsimplifyÕ it? One thing you can do is to compute the quotient and remainder upon dividing 5 into 21:FcarF%kBhc 21kBhc5kBhc = 4 + carF$kBhc1kBhc5EkBhcž—Our method for handling polynomials is similar, and is based on the analog of long division for integers. When confronted with a rational function likeDcarFrkBhcxopxEkBhc3kBhc  + xkBhcxopxEkBhc2kBhc  + 2CkBhc\Uwe will perform long division with polynomials to compute the quotient and remainder.BkBhc x - carFKkBhcxkBhcxopxEkBhc2kBhc  + 2AkBhc Thus@etnI–carFrkBhcxopxEkBhc3kBhc  + xkBhcxopxEkBhc2kBhc  + 2kBhcxkBhc  = etnI,kBhcxkBhcxkBhc  - etnIocarFKkBhcxkBhcxopxEkBhc2kBhc  + 2kBhcx?kBhc“ŒThe first integral on the right is trivial, while the second integral is also not too hard to integrate: we simply apply the substitution u kBhc=kBhc kBhcxopxEkBhc2kBhc  + 2.=kBhc%Here then is one last example.5etnIÄcarF kBhcxopxEkBhc2kBhc - 2 x + 1GrapkBhc x + 1Grap;kBhcxopxEkBhc2kBhc  + 1kBhcxikBhc³¬The rational function is in the correct form for partial fractions (the numerator has degree 2, while the denominator has degree 3), so we go directly to the decomposition.jcarF kBhcxopxEkBhc2kBhc - 2 x + 1GrapkBhc x + 1Grap;kBhcxopxEkBhc2kBhc  + 1kBhc  = carF(kBhcAkBhc x + 1kBhc  + carFQkBhcB x + CkBhcxopxEkBhc2kBhc  + 1kkBhc­¦Note that the numerators of each of the two new fractions have to have degree one less than their denominators. Next, multiply both sides by the original denominator.lkBhcxopxEkBhc2kBhc - 2 x + 1 = AGrap;kBhcxopxEkBhc2kBhc  + 1kBhc  + GrapkBhcB x + CGrapkBhc x + 1m kBhc2+We next make a series of substitutions for kBhcxkBhcIB in an effort to produce a set of simpler equations. Substituting kBhcxkBhc kBhc=kBhc kBhc-kBhc1 givesnkBhc4 = 2 AokBhc A = 2pkBhc Substituting kBhcxkBhc kBhc=kBhc 0 givesqkBhc1 = A + C = 2 + CrkBhc C = -1skBhc Substituting kBhcxkBhc kBhc=kBhc 1 givestkBhc 0 = 2 A + 2 GrapkBhc B + CkBhc = 4 + 2 B - 2ukBhc B = -1vkBhc Thusw etnIÄcarF kBhcxopxEkBhc2kBhc - 2 x + 1GrapkBhc x + 1Grap;kBhcxopxEkBhc2kBhc  + 1kBhcxkBhc  = etnILcarF(kBhc2kBhc x + 1kBhcxkBhc  + etnItcarFPkBhc -x - 1kBhcxopxEkBhc2kBhc  + 1kBhcxkBhc  = 2 etnILcarF(kBhc1kBhc x + 1kBhcxkBhc  - etnIocarFKkBhcxkBhcxopxEkBhc2kBhc  + 1kBhcxkBhc  - etnIocarFKkBhc1kBhcxopxEkBhc2kBhc  + 1kBhcxx kBhc = 2 lnVsbAkBhc x + 1kBhc  - carF$kBhc1kBhc2kBhc  lnVsbA;kBhcxopxEkBhc2kBhc  + 1kBhc  - tanopxEkBhc -1kBhc  x + C