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

Answer 1

William Abdalla

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

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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Answer 2

Elijah Abram

#x-ln(e^x+1)+C#

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-ln(e^x+1)+C#

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Answer 3

Christopher Allers

To integrate ( \frac{1}{e^x + 1} ) with respect to ( x ), you can use the substitution method. Let ( u = e^x + 1 ). Then, ( du = e^x dx ).

So, the integral becomes:

[ \int \frac{1}{u} du ]

This is a straightforward integral, which evaluates to:

[ \ln|u| + C ]

Substituting back ( u = e^x + 1 ), the final result is:

[ \ln|e^x + 1| + C ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.