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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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    @Zev: Or, more likely, you took `arcta$n$x` and made it into `\arctanx`, which doesn't exist...2012-05-19

2 Answers 2

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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