How to integrate $\int \frac{e^x dx}{1\,+\,e^{2x}}$ Ok, I give up, I have tried with $u$-substitution and integration by parts but I can't solve it. The integral is:

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

I have tried $u=e^x$, $u=e^{2x}$ and also integration by parts but I can't solve it. The result should be:

$$\arctan(e^x)$$
Use $u = e^x, du = e^x dx.$

Then you have:

$$\int \frac{du}{1 + u^2} \text{because} \space (e^x)^2 = e^{2x} $$

$$\arctan (u) + C$$

$$\arctan(e^x) + C$$
$$\newcommand{\arctg}{\operatorname{arctg}} \int{\frac{e^x dx}{1+e^{2x}}} = \int{\frac{du}{1+u^{2}}} = \arctg(u)+c = \arctg(e^x)+c$$