PostClass 7.4-solutions

I 2 a5 x b x 1 x 15 x x 15 x x5 3 dx

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Unformatted text preview: B= . 2 and 5. I = 3 ln 2 4 6. I = 2 A(5 − x) + B (x − 1) = , (x − 1)(5 − x) (x − 1)(5 − x) x=5 3 dx . +4 1. I = Explanation: To find A, B so that x=1 ex 5 4 ln 3 3 Explanation: Set u = ex + 4. Then du = ex dx = (u − 4) dx, while x=0 =⇒ u = 5, x = ln 2 =⇒ u = 6. 4 2 eakin (ace669) – PostClass 7.4 – morales – (56430) In this case, 2 By partial fractions, 6 I= 5 1 A Bx + C = +2 . 2 + 1) (x + 1)(x x+1 x +1 3 du . u(u − 4) To evaluate this last integral we need to use partial fractions: for then 1 1 1 1 . = − u(u − 4) 4 u−4 u To determine A, B, and C multiply through by (x + 1)(x2 + 1): for then 1 = A(x2 + 1) + (x + 1)(Bx + C ) = ( A + B ) x2 + ( B + C ) x + ( A + C ) , Thus I= 6 3 4 5 3 ln |u − 4| − ln |u| 4 = which after comparing coefficients gives 1 1 du − u−4 u = 6 5 A = −B , . 5 I= 1 2 Consequently, = I= 3 4 ln 1 1 − ln 3 5 = 35 ln . 43 1 1 . 2 1 x−1 dx −2 x+1 x +1 0 1 1 2 0 1 dx − x+1 1 + x2 0 003 A= Thus 6 u...
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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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