Math 231E. Fall 2013. Lecture 8B.Partial Fractions, Strategy for IntegrationWe know how to do integrals of the formZdxx − 3= ln |x − 3| + C,and we know how to do integrals of the formZdxx2+ 1= arctan(x) + C.We can also do integrals of the formZ3x dxx2+ 1=Z32duu=32ln |u| + C =32|x2+ 1| + C.All of these integrals are of what are called rational functions:Definition 7. A rational function is a functionf(x) =P (x)Q(x),where P and Q are polynomials.The idea behind the technique of partial fractions is that it allows us to compute the antideriva-tive of any rational function, after some work. The main idea behind it is that we write any rationalfunction as a sum of simpler rational functions that we know how to integrate, and more or lessthose three examples above are the prototypical examples that we need to consider.Said another way, partial fractions is mostly an algebraic technique that allows us to rewriteintegrands in a more useful form.We will first consider a couple of examples.Example 39. Let us consider the integralZdx(x − 1)(x − 2).Let us guess that we can write1(x − 1)(x − 2)=Ax − 1+Bx − 2for some numbers A, B. If we can do so, then we will be able to do these integrals. Let us simplifythat previous expression by putting everything over a common denominator1(x − 1)(x − 2)=A(x − 2)(x − 1)(x − 2)+B(x − 1)(x − 1)(x − 2),and then equating numerators1 = A(x − 2) + B(x − 1) = (A + B)x + (−2A − B).(We could also just have multiplied out the denominators, and obtained the same thing.)There are then two ways to solve for A, B. First, if we consider the equation(A + B)x + (−2A − B) = 1,and note that it must be true for all x, this gives the systemA + B = 0, −2A − B = 1,which has solution A = −1, B = 1. The second way is to consider the equation1 = A(x − 2) + B(x − 1),and note that it simplifies considerably for two special cases of x, namely x = 1, 2. If we plug inx = 1 we obtain1 = A(−1) = −A,so A = −1, and if we plug in x = 2 we obtain1 = B(1) = B,so B = 1.Either way, we have found that1(x − 1)(x − 2)=−1x − 1+1x − 2.ThusZdx(x − 1)(x − 2)= −ln |x − 1| + ln |x − 2| + C = lnx − 2x − 1+
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