If #sigma = int_0^1 (e^x)/(1+x) dx#, show that # int_0^1 (xe^x)/(1+x) dx = e - 1 - sigma# Hint: the cutest solution is often over-looked. ?

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Answer 1 · 2018-07-09T09:51:15

Please see below.

Here given that ,

#color(blue)(sigma=int_0^1 e^x/(1+x)dx...to(1)#

Let ,

#I=int_0^1 (x*e^x)/(1+x)dx#

#=>I=int_0^1 (x*e^xcolor(red)(+e^x-e^x))/(x+1)dx#

#=>I=int_0^1 (x*e^x+e^x)/(x+1)dx-int_0^1e^x/(1+x)dxto color (blue)([use (1)]#

#=>I=int_0^1 (e^xcancel((x+1)))/(cancel((x+1)))dx-color(blue)(sigma#

#=>I=int_0^1e^xdx-sigma#

#=>I=[e^x]_0^1 -sigma#

#=>I=e^1-e^0-sigma#

#=>I=e-1-sigma#