mt1-ans

# mt1-ans - Math 1B Sample Answers to First Midterm 1(10...

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Math 1B. Sample Answers to First Midterm 1. (10 points) (a). Find Z x tan - 1 x dx . Use integration by parts with u = tan - 1 x dv = x dx du = dx x 2 +1 v = x 2 2 : Z x tan - 1 x dx = x 2 2 tan - 1 x - 1 2 Z x 2 x 2 + 1 dx = x 2 2 tan - 1 x - 1 2 Z 1 - 1 x 2 + 1 dx = x 2 2 tan - 1 x - x 2 + 1 2 tan - 1 x + C . (b). Find Z x dx x 2 - 3 x 2 . Make the rationalizing substitution u = 3 x , viewed as the inverse substitution x = u 3 , dx = 3 u 2 du , and then do the substitution v = u 4 - 1 , dv = 4 u 3 du : Z x dx x 2 - 3 x 2 = Z u 3 · 3 u 2 du u 6 - u 2 = Z 3 u 3 du u 4 - 1 = 3 4 Z dv v = 3 4 ln | v | + C = 3 4 ln | u 4 - 1 | + C = 3 4 ln 3 x 4 - 1 + C . 2. (10 points) (a). Find Z x 2 - 1 x dx . Do the trigonometric substitution x = sec θ , dx = sec θ tan θ dθ . Here p x 2 - 1 = p sec 2 θ - 1 = p tan 2 θ = tan θ 1

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2 (or draw a diagram). We have Z x 2 - 1 x dx = Z tan θ sec θ tan θ dθ sec θ = Z tan 2 θ dθ = Z (sec 2 θ - 1) = tan θ - θ + C = p x 2 - 1 - sec - 1 x + C . (b). Find Z e - 2 0 dx x (ln x ) 2 . This is an improper integral, since the denominator 0 as x 0 + .
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