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

Question

My attempt. Rewriting it as $$ \frac{e^x e^x}{1+e^x} \text{ then } u = e^x \text{ and } du = e^x dx $$ Substitute to get $u/(1+u)$ and now integrate + Plug in $$ \int \frac{e^{2x}dx}{1+e^x} = \frac{e^{2x}}{2} + x $$

Answer 1

You made a mistake somewhere, possibly by integrating with respect to $x$ rather than $u$, in any case, $$\int\dfrac{u}{1+u}\,\mathrm{d}u=\int\dfrac{u+1}{1+u}\,\mathrm{d}u+\int\dfrac{-1}{1+u}\,\mathrm{d}u=u-\ln(1+u)+\mathrm{C}.$$