Unformatted text preview: + 1) = 1 √x 2x . 2. Evaluate the integral. ( Hint: Integration by parts, and then long division. ) Z x ln( x + 1) dx. Solution: Let u = ln( x + 1) ,dv = xdx,du = 1 x +1 dx,v = x 2 2 . Z x ln( x + 1) dx = ln( x + 1) x 2 2Z x 2 2 1 x + 1 dx = x 2 2 ln( x + 1)1 2 Z ± ( x1) + 1 x + 1 ² dx = x 2 2 ln( x + 1)1 2 ³ x 2 2x + ln( x + 1) ´ + C. 1...
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 Fall '06
 McAdam
 Calculus, Derivative, ln ln, Guoliang Wu

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