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This question already has an answer here:

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.

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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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    $\begingroup$ 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 $g$. $\endgroup$ – André Nicolas May 9 '15 at 15:32
  • $\begingroup$ When I try doing that, I end up with: (int)du/(u^2 + 1) $\endgroup$ – EuclidAteMyBreakfast May 9 '15 at 15:34
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    $\begingroup$ Yes, and $\int \frac{du}{1+u^2}=\arctan u+C$. Standard integral. If you do not recognize it, let $u=\tan t$. $\endgroup$ – André Nicolas May 9 '15 at 15:36
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    $\begingroup$ Nice nickname! :-) Mathematicians are so mean. Euclid, the meanest of all, goes around stealing breakfasts and making people cry. :-) $\endgroup$ – Giuseppe Negro May 9 '15 at 15:44
  • $\begingroup$ 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. $\endgroup$ – Martin Sleziak May 10 '15 at 9:16
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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 $$ \int \frac{1}{u^2 + 1}du=\arctan u+C. $$

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  • $\begingroup$ I'm not able to understand how (int)1/(u^2 + 1)du = arctanu + C $\endgroup$ – EuclidAteMyBreakfast May 9 '15 at 15:41
  • $\begingroup$ @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 $\arctan'(x)=\frac{1}{1+x^2}$ $\endgroup$ – Hasan Saad May 9 '15 at 15:52
  • $\begingroup$ I got it, thanks a lot :) $\endgroup$ – EuclidAteMyBreakfast May 9 '15 at 15:55

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