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 $u=1+e^x$
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 $C'=C+1$.