121616949-math.186 - x-r x-r x-s = A B x-As-Br x-r x-s That...

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172 Chapter 8 Techniques of Integration If x 2 + bx + c = ( x - r ) 2 then we have the special case we have already seen, that can be handled with a substitution. The other two cases require different approaches. If x 2 + bx + c = ( x - r )( x - s ), we have an integral of the form Z p ( x ) ( x - r )( x - s ) dx where p ( x ) is a polynomial. The first step is to make sure that p ( x ) has degree less than 2. EXAMPLE 8.18 Rewrite Z x 3 ( x - 2)( x + 3) dx in terms of an integral with a numerator that has degree less than 2. To do this we use “long division of polynomials” to discover that x 3 ( x - 2)( x + 3) = x 3 x 2 + x - 6 = x - 1 + 7 x - 6 x 2 + x - 6 = x - 1 + 7 x - 6 ( x - 2)( x + 3) , so Z x 3 ( x - 2)( x + 3) dx = Z x - 1 dx + Z 7 x - 6 ( x - 2)( x + 3) dx. The first integral is easy, so only the second requires some work. Now consider the following simple algebra of fractions: A x
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Unformatted text preview: ( x-r ) ( x-r )( x-s ) = ( A + B ) x-As-Br ( x-r )( x-s ) . That is, adding two fractions with constant numerator and denominators ( x-r ) and ( x-s ) produces a fraction with denominator ( x-r )( x-s ) and a polynomial of degree less than 2 for the numerator. We want to reverse this process: starting with a single fraction, we want to write it as a sum of two simpler fractions. An example should make it clear how to proceed. EXAMPLE 8.19 Evaluate Z x 3 ( x-2)( x + 3) dx . We start by writing 7 x-6 ( x-2)( x + 3) as the sum of two fractions. We want to end up with 7 x-6 ( x-2)( x + 3) = A x-2 + B x + 3 ....
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