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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:

$$\arctan(e^x)$$

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    $\begingroup$ $u = e^x$ works. Try it again. $\endgroup$ – Qiaochu Yuan Jun 24 '12 at 3:06
  • $\begingroup$ You should recognize $e^x = \frac{1}{2} \frac{d}{dx} (e^x)^2$. $\endgroup$ – Siminore Jun 24 '12 at 8:34
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Use $u = e^x, du = e^x dx.$

Then you have:

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

$$\arctan (u) + C$$

$$\arctan(e^x) + C$$

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As usual, recognizing patterns can make a great difference. Assuming we know $$\int\frac{dx}{1+x^2}=\arctan x+C\Longrightarrow \int\frac{d(f(x))}{1+f^2(x)}=\arctan(f(x))+C$$we have that, since $\,(e^x)'=e^x\,$ , then

$$\int\frac{e^x}{1+e^{2x}}\,dx=\int\frac{d(e^x)}{1+(e^x)^2}=\arctan e^x+C$$

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Let $u=e^x$. Then $du=e^x \,dx$ and $1+e^{2x}=1+u^2$. You should be able to finish from there. And don't forget the arbitrary constant of integration.

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You know the answer, so you can backtrack:

$$(\arctan e^x)'=\frac{(e^x)'}{1+(e^x)^2}=\frac{e^x}{1+e^{2x}}.$$

This should show you how substitution will work.

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

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    $\begingroup$ Mathematically a duplicate of two-year-old answers, and explained less well. $\endgroup$ – epimorphic Jan 12 '15 at 23:23

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