Question #8f70f

1 Answer
Deepak Pandey
Sep 16, 2017

#:.int1/(e^x+1)dx=x -ln (e^x+1) +c#
Where #c# is the constant of integration

Explanation:

#int1/(e^x+1)dx=int (1-e^x/(e^x+1))dx#
#=x-inte^x/(e^x+1)dx#
Now for #inte^x/(e^x+1)dx # substitute #e^x=t# then #e^xdx=dt#
Hence the integral becomes #intdt/(t+1)=ln (t+1)+c#
#:.inte^x/(e^x+1)dx= ln (e^x+1)+c#
#:.int1/(e^x+1)dx=x -ln (e^x+1)+c#

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