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hw8sol - Fall 2010 MAT 21B Solution to Paper Homework 8...

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Fall 2010, MAT 21B, Solution to Paper Homework # 8 Problem 1. Evaluate the integral Z 4 x 3 + 3 x 2 - 2 x + 1 ( x - 1) 2 ( x 2 + 2 x + 3) dx Solution 1. We first expand the expression 4 x 3 +3 x 2 - 2 x +1 ( x - 1) 2 ( x 2 +2 x +3) using the method of partial frac- tions. We look for expressions A , B , C , and D such that 4 x 3 + 3 x 2 - 2 x + 1 ( x - 1) 2 ( x 2 + 2 x + 3) = A x - 1 + B ( x - 1) 2 + Cx + D x 2 + 2 x + 3 Combining the terms on the right hand side and comparing coefficients of the numerator gives the following system of linear equations: A + C = 4 A + B - 2 C + D = 3 A + 2 B + C - 2 D = - 2 - 3 A + 3 B + D = 1 The unique solution is given by A = 2 , B = 1 , C = 2 , and D = 4 . Now, the integral simplifies to Z 4 x 3 + 3 x 2 - 2 x + 1 ( x - 1) 2 ( x 2 + 2 x + 3) dx = Z 2 x - 1 dx + Z 1 ( x - 1) 2 dx + Z 2 x + 4 x 2 + 2 x + 3 dx = Z 2 x - 1 dx + Z 1 ( x - 1) 2 dx + Z 2 x + 2 x 2 + 2 x + 3 dx + Z 2 x 2 + 2 x + 3 dx We can solve the first three integrals in the sum by the method of substitution, which yields Z 2 x - 1 dx + Z 1 ( x - 1) 2 dx + Z 2 x + 2 x 2 + 2 x + 3 dx + Z 2 x 2 + 2 x + 3 dx = 2 ln | x - 1 | - 1 x - 1 + ln | x 2 + 2 x + 3 | + C + Z 2 x 2 + 2 x + 3 dx The last integral can be simplified into a more familiar form: Z 2 x 2 + 2 x + 3 dx = Z 2 ( x + 1) 2 + 2 dx = Z 1 ( x +1 2 ) + 1 dx = 2 arctan x + 1 2 + C Finally, putting these results together yields the final answer Z 4 x 3 + 3 x 2 - 2 x + 1 ( x - 1) 2 ( x 2 + 2 x + 3) dx = 2 ln | x - 1 | - 1 x - 1 + ln | x 2 + 2 x + 3 | +
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