PostClass 7.4-solutions

0 points and so evaluate the integral 1 i 0 1 i i 1 dx

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Unformatted text preview: 4 3 ln 4 u B = −C , 1 0 x2 x dx +1 1 dx , +1 10.0 points and so Evaluate the integral 1 I= 0 1. I = I= 1 dx . (x + 1)(x2 + 1) π 1 ln(2) + 4 2 correct 1 1 ln(x + 1) − ln(x2 + 1) + tan−1 (x) 2 2 = Consequently, π 1 ln(8) − 2. I = 2 2 3. I = 1 π ln(8) − 4 2 1π − ln(2) 4. I = 22 5. I = I= Explanation: π 1 ln(2) + 4 2 004 10.0 points Evaluate the integral 2 I= 1π − ln(2) 42 π 1 ln(2) + 6. I = 2 2 (x + 1)2 1 ln 2 + 2 tan−1 (x) 4 x +1 1 1. I = 8 1 ln 4 5 2. I = 8 1 ln 16 5 1 dx . x3 + 4 x . 1 0 . 1 0 eakin (ace669) – PostClass 7.4 – morales – (56430) 3 3. I = 1 5 ln 4 2 2. I = x2 − 4x + 16 ln |x + 4| + C 4. I = 8 1 ln 8 5 3. I = x2 + 4x + 16 ln |x + 4| + C 1 5 5. I = ln correct 8 2 4. I = x2 + 4x + 16 ln |x + 4| + C 2 1 5 ln 16 2 5. I = x2 + 4x + 4 ln |x + 4| + C 2 6. I = x2 − 4x + 16 ln |x + 4| + C correct 2 6. I = Explanation: By partial fractions, 1 A Bx + C = +2 . x3 + 4 x x x +4 Explanation: After division we see that To determine A, B, and C multiply through by x3 + 4x: for then x2 − 16 16 x2 = + x+4 x+4 x+4 1 = A(x2 + 4) + x(Bx + C ) = x−4+ = (A + B )x2 + Cx + 4A , which afte...
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