Lecture6 - Lecture 6: MATH 138-W12-003 January 16, 2012 I)...

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Lecture 6: MATH 138-W12-003 January 16, 2012 I) Integration by Rational Functions by Partial Fractions, Section 7.4 We know very well that to sum two fractions that are polynomials in x we can get a common denominator. For example, 2 x - a + 1 x - b = 2( x - b ) + ( x - a ) ( x - a )( x - b ) = 3 x - (2 b + a ) ( x - a )( x - b ) What we will now learn how to do is do the reverse. First consider a rational function of the form, f ( x ) = P ( x ) Q ( x ) , where P ( x ) and Q ( x ) are polynomials and the degree of P ( x ) is less than that of Q ( x ) . First we now learn how to decompose this into separate terms that we can integrate using our standard methods. If the degree of the numerator is larger than that of the denominator than we must use long division. This reduces our rational polynomial into a sum of a polynomial and a rational polynomial where the numerator has a smaller degree than the denominator, P ( x ) Q ( x ) = polynomial + R ( x ) Q ( x ) , where degree R ( x ) is less than the degree of Q ( x ) . Below is an example of how you can do this using long division and that this yields to a simpler form of the integrand that we can evaluate more readily. Example: Evaluate, Z x 3 + x x - 1 dx. Step 1): First observe that the numerator has a larger degree than the denominator so we must use long division to reduce it to a polynomial and a proper rational function. x 2 + x + 2 x - 1 ) x 3 + x - x 3 + x 2 x 2 + x - x 2 + x 2 x - 2 x + 2 2 Step 2): Reduce the denominator Q ( x ) to polynomials we can factor and then expand. In this case it is not necessary since we simply have x - 1 that we can integrate. Step 3): The result is something we can integrate so we substitute this into the integrand and eval- uate the integral using our known techniques, Z x 3 + x x - 1 dx = Z ± x 2 + x + 2 + 2 x - 1 ² dx, = 1 3 x 3 + 1 2 x 2 + 2 x + 2 ln | x - 1 | + C. 1
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In general, after using long division we must use partial fractions to rewrite the rational func- tion as separate terms of the form, A ( ax + b ) n , or Ax + B ( ax 2 + bx + c ) n . Each of which can be integrated using a substitution. It can be shown that we can always do this decomposition. Instead of proving this we focus on learning how this can be done. The textbook deals with four different cases. We only learn the first three. The interested reader is directed to the textbook for the fourth case. 1 Case I: The denominator
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This note was uploaded on 02/14/2012 for the course MATH 138 taught by Professor Anoymous during the Fall '07 term at Waterloo.

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Lecture6 - Lecture 6: MATH 138-W12-003 January 16, 2012 I)...

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