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Integrate the following: $$\int\frac{e^{\arctan(x)}}{1+x^2}\,dx$$

let $u = \arctan(x)$ , then $du = \frac{1}{1+x^2} dx$

$$\int e^u du$$ $$e^u + C$$ $$e^{\arctan(x)} + C$$

I am not sure if I did this right, this was a question on the final that I had today. Any help checking is appreciated.

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    [Looks good](http://www.google.com/search?q=thumbs+up&hl=en&client=firefox-a&hs=KLH&rls=org.mozilla:en-US:official&prmd=imvns&source=lnms&tbm=isch&ei=vxq3T_CHDIa29QSiwa2xCg&sa=X&oi=mode_link&ct=mode&cd=2&ved=0CEUQ_AUoAQ&biw=1355&bih=1235&sei=xRq3T84ogaTwBN-3vc0K).2012-05-19
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    @Zev: $\LaTeX$ is perfectly familiar with the arctangent function; you can use `\arctan`, no need to use `\operatorname{arctan}`.2012-05-19
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    @Arturo: Thanks for correcting it! I'd thought I'd gotten a red MathJax error when I tried it, but perhaps I just misspelled "arctan".2012-05-19
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    @Zev: Or, more likely, you took `arctanx` and made it into `\arctanx`, which doesn't exist...2012-05-19

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Well done. But you do not need me to check it. You can check it for yourself by differentiating. This is the case for any indefinite integral.

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He has a point, take the derivative of $e^{\arctan(x)}$

which is done by $e^u \cdot du/dx$

So you have $e^{\arctan(x)}/(1+x^2)$ which was your original integral