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

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29
Nov 19, 2016

Answer:

Multiply top and bottom by #e^-x#.

Explanation:

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

Divide numerator and denominator by #e^x#:

#=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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23
Apr 11, 2017

#int frac{dx}{e^x+1}=int frac{e^x+1-e^x}{e^x+1}dx=int (1-frac{e^x}{e^x+1})dx=x-ln(e^x+1)#

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11
Jan 29, 2017

Answer:

Use the substitution #u=e^x# and partial fraction decomposition.

Explanation:

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

Apply the substitution #u=e^x#:

#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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mason m Share
Mar 25, 2017

Let #e^(x/2)=tantheta#. Then #1/2e^(x/2)dx=sec^2thetad theta#.

#intdx/(e^x+1)=2int(1/2e^(x/2)dx)/(e^(x/2)(e^x+1))=2int(sec^2thetad theta)/(tantheta(sec^2theta))=2intcostheta/sinthetad theta#

#=2lnabssintheta#

From #tantheta=e^(x/2)# draw a right triangle to see that #sintheta=e^(x/2)/sqrt(e^x+1)#:

#=2lnabs(e^(x/2)/sqrt(e^x+1))=lnabs(e^x/(e^x+1))=x-lnabs(e^x+1)+C#

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