A02 Soln - Math 138 Assignment 2 Solutions 1. Z x 3 + 1 x...

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Unformatted text preview: Math 138 Assignment 2 Solutions 1. Z x 3 + 1 x 2- 2 x + 5 dx Use long division to get x 3 + 1 x 2- 2 x + 5 = x + 2- x + 9 x 2- 2 x + 5 The first two terms are easy to deal with so we focus on the third term. Complete the square to get x + 9 x 2- 2 x + 5 = x + 9 ( x- 1) 2 + 4 then let u = x- 1 ,du = dx so we have Z x + 9 ( x- 1) 2 + 4 dx = Z u + 10 u 2 + 4 du = Z u u 2 + 4 du + 10 Z 1 u 2 + 4 du For Z u u 2 + 4 du let s = u 2 + 4 so ds = 2 udu therefore Z u u 2 + 4 du = 1 2 Z 1 s ds = 1 2 ln( s ) = 1 2 ln( u 2 + 4) = 1 2 ln( x 2- 2 x + 5) . Note s > 0 so we don’t need the absolute value signs in the ln function. For Z 1 u 2 + 4 du we simply have Z 1 u 2 + 4 du = Z 1 4( u 2 4 + 1) du = 1 4 Z 1 u 2 2 + 1 du = 1 2 arctan u 2 = 1 2 arctan x- 1 2 . Therefore Z x + 9 ( x- 1) 2 + 4 dx = Z u u 2 + 4 du + 10 Z 1 u 2 + 4 du = 1 2 ln( x 2- 2 x + 5) + 5 arctan x- 1 2 . Our full solutions is thus Z x 3 + 1 x 2- 2 x + 5 dx = Z x +2- x + 9 x 2- 2 x + 5 dx = x 2 2 +2 x- 1 2 ln( x 2- 2 x +5)- 5 arctan...
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This note was uploaded on 02/14/2012 for the course MATH 138 taught by Professor Anoymous during the Fall '07 term at Waterloo.

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A02 Soln - Math 138 Assignment 2 Solutions 1. Z x 3 + 1 x...

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