Friday, May 20, 2016

calculus - Evaluate intinfty0frac1sqrtx(1+ex)mathrmdx


I would like to evaluate: 01x(1+ex)dx Given that I can't find 1x(1+ex)dx, a substitution is needed: I tried u=x(1+ex) and u=1x(1+ex) but I could not get rid of x in the new integral.... Do you have ideas of substitution?


Answer



01x(1+ex)dx=2011+ex2dx=20(1+ex2)1/2ex2/2dx=20k=0(14)k(2kk)e(2k+1)x2/2dx=k=0(14)k(2kk)2π2k+1


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