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Unformatted text preview: MATH 118, LECTURES 7 & 8: PARTIAL FRACTIONS 1 Partial Fractions Consider being asked to solve integraldisplay x ( x 1)( x + 1)( x + 3) dx. We may be tempted to try a variety of different substitutions; however, there is a significantly simpler method by which this integral (and those like it) can be solved. The method is called partial fraction decomposition . What we notice here is that if any one of the three components on the bottom appeared by itself, we would be able to integrate immediately. For instance, we can evaluate integraldisplay 1 x 1 dx = ln  x 1  + C. That was awfully easy. It would make our lives significantly easier if we could separate the term 1 / ( x 1)( x + 1)( x + 3) into three fractions of the form, say, A/ ( x 1), B/ ( x + 1), and C/ ( x + 3). There is nothing telling us that we are not allowed to do this, so lets try it. We set x ( x 1)( x + 1)( x + 3) = A x 1 + B x + 1 + C x + 3 . (1) So how do we solve for the constants A , B and C ? We will go over the general methods for doing this in a moment, but for our purposes here, lets just notice that we can multiply across by the denominator on the lefthand side to get x = A ( x + 1)( x + 3) + B ( x 1)( x + 3) + C ( x 1)( x + 1) . We notice that, in order for (2) to be satisfied, it must be satisfied for all x . This means that we can select any value of x we want to solve for the constants A , B and C ! 1 Which values of x should we choose. We notice the brackets on the righthand side have zeroes at the values x = 1, x = 1, and x = 3. It will simplify our algebra to use these values. Setting x = 1 we obtain 1 = A (2)(6) which implies A = 1 8 . Setting x = 1 we obtain 1 = B ( 2)(2) which implies B = 1 4 . Setting x = 3 we obtain 3 = C ( 4)( 2) which implies C = 3 8 . That was not nearly as painful as it could have been, but lets keep in mind what we have actually done. This means that x ( x 1)( x + 1)( x + 3) = 1 8( x 1) + 1 4( x + 4) 3 8( x + 3) (Notice that we can check to see if this answer is validi.e. whether the process is validby simply finding a common denominator. The process is tedious, but we might as well go through it once to convince ourselves that our method works: 1 8( x 1) + 1 4( x + 1) 3 8( x + 3) = 1 4 bracketleftbigg 1 2( x 1) + 1 x + 1 3 2( x + 3) bracketrightbigg = 1 4 bracketleftbigg ( x + 1)( x + 3) + 2( x 1)( x + 3) 3( x + 1)( x 1) 2( x 1)( x + 1)( x + 3) bracketrightbigg = 1 8 bracketleftbigg x 2 + 4 x + 3 + 2 x 2 + 4 x 6 3 x 2 + 3 ( x 1)( x + 1)( x + 3) bracketrightbigg = x ( x 1)( x + 1)( x + 3) . So our partial fraction separation checks out!) The only remaining step is to integrate. If we have done things correctly, this should be the easiest part of the whole problem!...
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This note was uploaded on 04/14/2009 for the course ENGINEERIN MATH 118 taught by Professor Soares during the Spring '09 term at Waterloo.
 Spring '09
 Soares

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