This preview shows page 1. Sign up to view the full content.
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 dx = x 2 2 ln( x + 1)1 2 x 2 2x + ln( x + 1) + C. 1...
View
Full
Document
This note was uploaded on 12/03/2010 for the course M 408c taught by Professor Mcadam during the Fall '06 term at University of Texas at Austin.
 Fall '06
 McAdam
 Calculus

Click to edit the document details