# How do you integrate # (1/(e^x+1))dx #?

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Let

From

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Note that:

#1/(e^x+1) = e^(-x)/(1+e^(-x)) = -d/(dx) ln (1+e^(-x))#

So:

#int \ 1/(e^x+1) \ dx = -ln(1+e^(-x))+C#

If you prefer, note that:

#-ln(1+e^(-x)) = -ln((e^x+1)/(e^x))#

#color(white)(-ln(1+e^(-x))) = ln(e^x)-ln(e^x+1)#

#color(white)(-ln(1+e^(-x))) = x-ln(e^x+1)#

So the integral can be expressed as:

#int \ 1/(e^x+1) \ dx = x-ln(e^x+1)+C#

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