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MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� � � Lecture 29 18.01 Fall 2006 Lecture 29: Partial Fractions We continue the discussion we started last lecture about integrating rational functions. We defined a rational function as the ratio of two polynomials: P (x) Q(x) We looked at the example 1 3 x − 1+ x + 2 dx = ln |x − 1| + 3 ln |x + 2| + c That same problem can be disguised: 1+3 =(x + 2) + 3(x − 1)= 4x − 1 x − 1 x + 2 (x − 1)(x + 2) x2 + x − 2 which leaves us to integrate this: � 4x − 1 dx = ??? x2 + x − 2 P (x)Goal: we want to figure out a systematic way to split into simpler pieces. Q(x) First, we factor the denominator Q(x). 4x − 1= 4x − 1= A + B x2 + x − 2 (x − 1)(x + 2) x − 1 x + 2 There’s a slow way to find A and B. You can clear the denominator by multiplying through by (x − 1)(x + 2): (4x − 1) = A(x + 2) + B(x − 1) From this, you find 4 = A + B and − 1 = 2A − B You can then solve these simultaneous linear equations for A and B. This approach can take a very long time if you’re working with 3, 4, or more variables. There’s a faster way, which we call the “cover-up method”. Multiply both sides by (x − 1): 4x − 1 B x + 2 = A + x + 2 (x − 1) Set x = 1 to make the B term drop out: 4 − 1= A1 + 2 A = 1 1Lecture 29 18.01 Fall 2006 The fastest way is to do this in your head or physically cover up the struck-through terms. For instance, to evaluate B: 4x − 1 A� B (x − 1)����x − 1 (x + 2)(x + 2) = ��+ ����Implicitly, we are multiplying by (x + 2) and setting x = −2. This gives us 4(−2) − 1= B = B = 3 −2 − 1 ⇒ What we’ve described so far works when Q(x) factors completely into distinct factors and the degree of P is less than the degree of Q. If the factors of Q repeat, we use a slightly different approach. For example: x2 + 2 A B C = + +(x − 1)2(x + 2) x − 1 (x − 1)2 x + 2 Use the cover-up method on the highest degree term in (x − 1). x2 + 1 12 + 2 = B + [stuff](x − 1)2 = = B = B = 1 x + 2 ⇒ 1 + 2 ⇒ Implicitly, we multiplied by (x − 1)2, then took the limit as x → 1. C can also be evaluated by the cover-up method. Set x = −2 to get x2 + 2 2 = C + [stuff](x + 2) = (−2)2 + 2 = C = C =2 (x − 1) ⇒ (−2 − 1)2 ⇒ 3 This yields x2 + 2 A 1 2/3 = + +(x − 1)2(x + 2) x − 1 (x − 1)2 x + 2 Cover-up can’t be used to evaluate A. Instead, plug in an easy value of x: x = 0. 2 A 1 1 1 (−1)2(2) = −1+ 1 + 3 = ⇒ 1 = 1 + 3 − A = ⇒ A =3 Now we have a complete answer: x2 + 2 1 1 2 = + +(x − 1)2(x + 2) 3(x − 1) (x − 1)2 3(x + 2) Not all polynomials factor completely (without resorting to using complex numbers). For exam-ple: 1 A1 B1x + C1 = +(x2 + 1)(x − 1) x − 1 x2 + 1 We find A1, as usual, by the cover-up method. 1 1 = A1 = A1 = 12 + 1 ⇒ 2 2� � � � � � Lecture 29 18.01 Fall 2006 Now, we have 1 1/2 B1x + C1 = +(x2 + 1)(x − 1) x − 1 x2 + 1 Plug in x = 0. 1 1 C1 1 1(−1) = − 2+1 = ⇒ C1 = − 2 Now, plug in any value other than x = 0, 1. For example, let’s use x = −1. 1 =1/2+ B1(−1) − 1/2= 0 = − B1 − 1/2= B1 = − 1 2(−2) −2 2 ⇒ 2 ⇒ 2 Alternatively, you can multiply out to clear the denominators (not done here). Let’s try to integrate this function, now. dx 1 dx 1 x dx 1 dx (x2 + 1)(x − 1) = 2 x − 1 − 2 x2 + 1 − 2 x2 + 1 1 1 1 = 2ln |x − 1| − 4 ln | x 2 + 1 | − 2 tan−1 x + c What if we’re faced with something that looks like this? dx (x − 1)10 This is actually quite simple to integrate: dx 1 (x − 1)10 = − 9(x − 1)−9 + c What about this? � dx (x2 + 1)10 Here, we would use trig substitution: x = tan u and dx = sec2 udu and the trig identity tan2 u + 1 = sec2 u to get � � sec2 u du = cos18 u du (sec2 u)10 From here, we can evaluate this integral using the methods we introduced two lectures ago.


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MIT 18 01 - Partial Fractions

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