Find $\int{\frac{e^{2x}}{1+e^x}}dx$

I did the following substitution: $u=e^x\Rightarrow du=e^x dx\Rightarrow \int{\frac{u}{1+u}du}=\int1du-\int\frac{1}{u+1}du=u-\ln{u}=e^x-\ln{(e^x+1)}+C $

I know this is the correct answer, however my initial approach was as follows: $u=1+e^x\Rightarrow du=e^xdx $ and $e^x=u-1\Rightarrow \int\frac{u-1}{u}=\int1du-\int\frac{1}{u}du=u-\ln u=1+e^x-\ln (1+e^x)+C$. Are these two answers equivalent due to the arbitrariness of the constant? Or is there an error in what I did in my substitution of $u=1+e^x$

  • 4
    $\begingroup$ Yes, the constant $C$ completely absorbs the $1$... $\endgroup$
    – abiessu
    Aug 4, 2021 at 18:12
  • $\begingroup$ How did you go from $e^{2 x}$ to just $u$ in the numerator?? $\endgroup$ Aug 4, 2021 at 19:28
  • $\begingroup$ @DavidG.Stork since $e^2x = (e^x)^2$ then one of those gets taken care of with the $du = e^x dx$ and the other $e^x$ can be replaced with $u$ $\endgroup$
    – mmmmmm
    Aug 5, 2021 at 19:34

1 Answer 1


They are the same answer, for the reason that you have mentioned:$$1+e^x-\ln(1+e^x)+C=e^x-\ln(1+e^x)+C'$$if we take $C'=C+1$.


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