Quiz 1 Solution - Solutions to Quiz 1 1 2x3 dx 1 x2 Solution Substitute u = 1 x2 so du = 2x dx Then u1 1 du = u du u u 2 2 1\/2 3\/2 C = u3\/2 2u1\/2 C = 1

Quiz 1 Solution - Solutions to Quiz 1 1 2x3 dx 1 x2...

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Solutions to Quiz 1 1. 2 x 3 1 + x 2 dx Solution : Substitute u = 1 + x 2 , so du = 2 x dx . Then 2 x 3 1 + x 2 dx = u - 1 u du = u - 1 u du = 2 3 u 3 / 2 - 2 u 1 / 2 + C = 2 3 ( 1 + x 2 ) 3 / 2 - 2 ( 1 + x 2 ) 1 / 2 + C. 2. ln x x 2 dx Solution : Integration by parts: u = ln x v = - 1 x du = 1 x dx dv = 1 x 2 dx So ln x x 2 dx = - ln x x - - 1 x · 1 x dx = - ln x x + x - 2 dx = - ln x x - 1 x + C. 3. sec 2 ( x ) dx Solution : Substitute u = x , so du = 1 2 x dx = 1 2 u dx . Then sec 2 ( x ) dx = sec 2 ( u )2 u du. Integration by parts gives:

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