How do you integrate #1/(e^x(e^x+1))# using partial fractions?

1 Answer
Andrea S.
Mar 4, 2017

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

Explanation:

Rather than using partial fractions directly we can prepare the function for a substitution. As the substitution will have to remove the exponential, let's bring it to the numerator noting that: #1/e^x = e^-x#:

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

Substitute now #e^-x = t#, #dt =-e^-xdx#

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

Now separate in partial fractions simply adding and subtracting #1# to the numerator:

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

This are regular integrals we can solve straight away:

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

and undoing the substitution:

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

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