# 8.3.17-eg - A C = 0-3 A B-4 C = 3 2 A-B 4 C =-1 whose...

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Example Evaluate Z 3 x - 1 ( x - 2)( x 2 - 3 x + 2) dx . 1. Factor the integrand and express as a sum of partial fractions. Notice that ( x - 2) is a repeated linear factor of the denominator: 3 x - 1 ( x - 2)( x 2 - 3 x + 2) = 3 x - 1 ( x - 2)( x - 1)( x - 2) = 3 x - 1 ( x - 2) 2 ( x - 1) so that partial fraction expansion is of the form: 3 x - 1 ( x - 2) 2 ( x - 1) = A x - 2 + B ( x - 2) 2 + C x - 1 . Multiplying through by the least common denominator ( x - 2) 2 ( x - 1) leads to the equation 3 x - 1 = A ( x - 2)( x - 1) + B ( x - 1) + C ( x - 2) 2 = Ax 2 - 3 Ax + 2 A + Bx - B + Cx 2 - 4 Cx + 4 C. Setting coefficients equal, we find
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Unformatted text preview: A + C = 0-3 A + B-4 C = 3 2 A-B + 4 C =-1 , whose solution is A =-2 ,B = 5 ,C = 2. 2. Integrate. Rewrite the integral as Z 3 x-1 ( x-2)( x 2-3 x + 2) dx = Z-2 x-2 + 5 ( x-2) 2 + 2 x-1 dx. Using substitution in the second term ( u = x-2), we obtain Z 3 x-1 ( x-2)( x 2-3 x + 2) dx =-2 ln | x-2 | -5 x-2 + 2 ln | x-1 | + C = 2 ln ± ± ± ± x-1 x-2 ± ± ± ±-5 x-2 + C....
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