PostClass 7.4-solutions

4 16 x4 thus i x4 16 x4 dx consequently thus i 2 1 4

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Unformatted text preview: r comparing coefficients gives A = −B , C = 0, 1 A= . 4 16 . x+4 Thus I= x−4+ 16 x+4 dx . Consequently, Thus I= 2 1 4 1 x 1 dx −2 x x +4 I= 1 1 ln(x) − ln(x2 + 4) 4 2 = = x2 1 ln 2 8 x +4 2 with C an arbitrary constant. 1 006 2 1 x2 − 4x + 16 ln |x + 4| + C 2 . Determine the integral Consequently, I= I= 5 1 ln 8 2 10.0 points dx . (x − 1)(x + 3) . 1. I = 005 2. I = 1 ln 4 x−1 x+3 3. I = 1 ln 2 2x + 2 (x − 1)(x + 3) 10.0 points Determine the integral I= 1 ln (|(x − 1)(x + 3)|) + C 2 x2 dx . x+4 1. I = x2 − 4x + 4 ln |x + 4| + C + C correct 4. I = ln (|(x − 1)(x + 3)|) + C +C eakin (ace669) – PostClass 7.4 – morales – (56430) 5. I = 1 ln 4 x+3 x−1 5. Area = 6 + +C Explanation: By partial fractions, 1 A B = + , (x − 1)(x + 3) x−1 x+3 4 77 sq.units ln 42 Explanation: The graph of f is an even function opening upwards, while the graph of g is a horizontal straight line. Thus the required area is similar to the shaded region in the figure below...
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This homework help was uploaded on 03/21/2014 for the course M 408S taught by Professor Stepp during the Spring '11 term at University of Texas at Austin.

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