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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. – David Mitra May 19 '12 at 4:00
@Zev: $\LaTeX$ is perfectly familiar with the arctangent function; you can use \arctan, no need to use \operatorname{arctan}. – Arturo Magidin May 19 '12 at 4:03
@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". – Zev Chonoles May 19 '12 at 4:03
@Zev: Or, more likely, you took arctanx and made it into \arctanx, which doesn't exist... – Arturo Magidin May 19 '12 at 4:05
up vote 9 down vote accepted

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

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