Question #7217e

1 Answer
Roy E.
Oct 14, 2017

#=cx+(e/a) ln (e^(ax-b)+1)+k#

Explanation:

#int c+e/(1+e^(-ax+b))dx#
#=cx+e int 1/(1+e^(-ax+b)dx#
Multiply top and bottom by #e^(ax-b)#
#=cx+e int (e^(ax-b))/(e^(ax-b)+1)dx#
The integrand is now nearly of the standard form #(f prime(x))/f(x)#
#=cx+(e/a) ln (e^(ax-b)+1)+k#

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