Math 231E. Fall 2013. Lecture 8C.Partial Fractions IIExample 40. We can actually generalize this example considerably. Let us now consider theintegralZ(αx + β) dx(x − 1)(x −2),where α, β are any numbers. We can then try the same decomposition:(αx + β)(x − 1)(x −2)=Ax − 1+Bx − 2,which becomesαx + β = (A + B)x + (−2A − B).This gives the two equationsA + B = α, −2A − B = β.Adding the two equations gives−A = α + β, or, A = −α − β,and then we have B = α − A = 2α + β. Thus we have(αx + β)(x − 1)(x −2)=−α − βx − 1+2α + βx − 2,and thusZ(αx + β) dx(x − 1)(x −2)= −(α + β) ln |x − 1| + (2α + β) ln |x − 2| + C.This technique will always work as long as the degree of the numerator is less than the degreeof the denominator, i.e. as long as it is a proper fraction. What if it is not?Example 41. We now considerZx3+ 4 dxx2− 3x + 2.We first have to do long division to make this a proper fraction. We saw how to write long divisionas a division problem earlier in the course, so let us now repeat this technique, but in a differentway.Let us write P (x) = x3+ 4 and Q(x) = x2− 3x + 2. We see then that to match the leadingorder terms, we have to multiply Q(x) by x, so let us see what happens:P (x) −xQ(x) = x3+ 4 − x(x2− 3x + 2) = x3+ 4 − x3+ 3x2− 2x = 3x2− 2x + 4.To remove the now leading order term, we have to multiply by 3, so we haveP (x) −xQ(x) − 3Q(x) = 3x2− 2x + 4 −3(x2− 3x + 2) = 3x2− 2x + 4 −3x2+ 9x − 6 = 7x −2.Thus we have shown thatP (x) −(x + 3)Q(x) = 7x − 2.Now that the remainder is of a lower degree that Q(x), we cannot proceed any further and we stophere. Thus we haveP (x)Q(x)− (x + 3) =7x − 2Q(x),orx3+ 4 dxx2− 3x + 2= x + 3 +7x − 2x2− 3x + 2.Of course, we know how to integrate x + 3, and we have already done the quotient in the generalcase above, with α = 7 and β = −2, so we havex3+ 4 dxx2− 3x + 2=x22+ 3x − 5 ln |x − 1| + 12 ln |x − 2|+ C.Example 42. This technique of partial fractions can extend to multiple terms in the denominator.For example, if we want to integrate a function of the formZ2x + 3 dx(x − 1)(x −2)(x −3)(x − 4),then we first write2x + 3(x − 1)(x −2)(x −3)(x − 4)=Ax − 1+Bx − 2+Cx − 3+Dx − 4.Expanding this out gives2x+3 = A(x−2)(x−3)(x−4)+B(x−1)(x−3)(x−4)+C(x−1)(x−2)(x−4)+D(x−1)(x−2)(x−3).Now, we could expand the right-hand side out and equate coefficients, but this is a case whereplugging in is more efficient. If we plug in the four numbers x = 1, 2, 3, 4 into this equation, weobtain the system5 = A(−1)(−2)(−3) = −6A,7 = B(1)(−1)(−2) = 2B,9 = C(2)(1)(−1) = −2C,11 = D(3)(2)(1) = 6D.This gives usA = −56, B =72, C = −92, D =116.Putting this all together givesZ2x + 3 dx(x − 1)(x −2)(x −3)(x − 4)= −56ln |x − 1| +72ln |x − 2| −92ln |x − 3| +116ln |x − 4| + C.Another case is if we have a repeated root in the denominator, along the lines of the followingexample:Example 43. Let us considerZ(2x + 1) dx(x − 1)2(x − 2).It turns out for the partial fractions we need to choose three terms (since there are three factorsdown below), and we do this by choosing as follows:2x + 1(x − 1)2(x − 2)=Ax − 1+B(x − 1)2+Cx − 2.Multiplying out gives2x + 1 = A(x −1)(x −2) + B(x − 2) + C(x − 1)2.The trick we used before of plugging in special values will not quite work here (although a refinementof this will work later). Let us multiply out the right-hand side and obtain2x+1 = A(x2−3x+2)+B(x−2)+C(x2−2x+1) = (A+C)x2+(−3A+B −2C)x+(2A−2B +C),which gives us the systemA + C = 0,−3A + B − 2C = 2,2A − 2B + C = 1.The first equation tells us that C = −A, so this simplifies the last two equations to−A + B = 2, A − 2B = 1,which givesA = −5, B = −3, C = 5,and thus2x + 1(x − 1)2(x − 2)=−5x − 1+−3(x − 1)2+5x − 2.ThusZ2x + 1(x − 1)2(x − 2)= −5 ln |x − 1|+3x − 1+ 5 ln |x − 2| + C.In general, whenever we have (x−a) to a power larger than one in the denominator, say (x−a)p,then we must have the p termsA1x − a+A2(x − a)2+ ··· +Ap(x − a)pin the partial fraction expansion.Example 44. Finally, if we ever have terms in the denominator that are quadratics with no realroots, then what we do is not factor the quadratic. However, since we will have two powers of x inthe denominator, we need two parameters in the numerator, so we always choose as followsx + 1(x − 2)(x2+ 2)=Ax − 2+Bx + Cx2+ 2.Writing this out givesx + 1 = A(x2+ 2) + (Bx + C)(x − 2) = (A + B)x2+ (−2B + C)x + (2A − 2C),which givesA + B = 0,−2B + C = 1,2A − 2C = 1.We can solve this asA = 1/2, B = −1/2, C = 0,sox + 1(x − 2)(x2+ 2)=1/2x − 2−12xx2+
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