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

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Multiply top and bottom by

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#int1/(e^x+1)dx#

Divide numerator and denominator by

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

Rewrite so that the numerator is the derivative of the denominator:

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

Integrate directly:

#=-ln|1+e^-x|+C#

We can leave the answer as is, or simplify further:

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

#=ln|e^x/(e^x+1)|+C#

#=x-ln|e^x+1|+C#

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Use the substitution

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#I=int1/(e^x+1)dx#

Apply the substitution

#I=int1/(u+1)((du)/u)#

#color(white)(I)=int1/(u(u+1))du#

Apply partial fraction decomposition:

#I=int(1/u-1/(u+1))du#

Integrate term by term:

#I=ln|u|-ln|u+1|+C#

Reverse the earlier substitution:

#I=x-ln|e^x+1|+C#

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Let

From

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