# Integration of $\int \frac{e^x}{e^{2x} + 1}dx$ [duplicate]

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I came across this question and I was unable to solve it. I know a bit about integrating linear functions, but I don't know how to integrate when two functions are divided. Please explain. I'm new to calculus.

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

Thanks in advance.

## marked as duplicate by Martin Sleziak, N. F. Taussig, J.-E. Pin, Jonas Meyer, user147263 May 10 '15 at 16:42

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• Let $u=e^x$. Integrate. You should end up with $\arctan u+C$. Note that there is no general procedure for finding the integral of \frac{f(x)}}{g(x)}$even when we know everything about the integrals of$f$and$g$. – André Nicolas May 9 '15 at 15:32 • When I try doing that, I end up with: (int)du/(u^2 + 1) – EuclidAteMyBreakfast May 9 '15 at 15:34 • Yes, and$\int \frac{du}{1+u^2}=\arctan u+C$. Standard integral. If you do not recognize it, let$u=\tan t$. – André Nicolas May 9 '15 at 15:36 • Nice nickname! :-) Mathematicians are so mean. Euclid, the meanest of all, goes around stealing breakfasts and making people cry. :-) – Giuseppe Negro May 9 '15 at 15:44 • Please, try to make the titles of your questions more informative. E.g., Why does$a<b$imply$a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Martin Sleziak May 10 '15 at 9:16 ## 1 Answer You may write $$\int \frac{e^x}{e^{2x} + 1}dx=\int \frac{d(e^x)}{(e^{x})^2 + 1}=\arctan (e^x)+C$$ since $$\int \frac{1}{u^2 + 1}du=\arctan u+C.$$ • I'm not able to understand how (int)1/(u^2 + 1)du = arctanu + C – EuclidAteMyBreakfast May 9 '15 at 15:41 • @EuclidAteMyBreakfast, it's a rule you should probably know.$(arctan u)'=\frac{1}{1+u^2}$. As for proving that, use$\tan(\arctan(x))=x$. Thus by the chain rule$\arctan'(x)(tan^2(\arctan(x))+1)=1$, that is$\arctan'(x)=\frac{1}{1+x^2}\$ – Hasan Saad May 9 '15 at 15:52
• I got it, thanks a lot :) – EuclidAteMyBreakfast May 9 '15 at 15:55