A02 Soln

# A02 Soln - Math 138 Assignment 2 Solutions 1 x3 1 dx x2 2x...

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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 u du 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 .

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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 x - 1 2 + C. 2. (a) Z 1 1 - x 2 dx Use partial fractions to get 1 1 - x 2 = 1 (1 - x )(1 + x ) = 1 2 1 1 - x + 1 1 + x Therefore Z 1 1 - x 2 dx = 1 2 ( - ln | 1 - x | + ln | 1 + x | ) + C = 1 2 ln 1 + x 1 - x + C (b) Z sec( x ) dx
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