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Consider:

$$\int \frac{1}{1+\arctan(x)}dx$$

I have attempted with making $x=\tan(u)$ and $\frac{dx}{du}=\sec^2(u)$

then ended up with:

$$\int \frac{\sec^2(u)}{1+u}du$$

$$\int \frac{1+\tan^2(u)}{1+u}du$$

$$\int \frac{1}{1+u}du+\int \frac{\tan^2(u)}{1+u}du$$

$$\ln\left|1+\arctan(x)\right|+\int \frac{\tan^2(u)}{1+u}du$$

And I do not know what to do from there on.

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    $\begingroup$ any reason to believe that a simple closed form exists? $\endgroup$ – tired Mar 8 '16 at 10:40
  • $\begingroup$ integrals.wolfram.com/index.jsp?expr=tan%28x%29^2%2F%281%2Bx%29&random=false $\endgroup$ – tired Mar 8 '16 at 10:43
  • $\begingroup$ When graphed on calculator, appears to give a similar type to $1/x$ $\endgroup$ – Kenny Guy Mar 9 '16 at 10:29
  • $\begingroup$ u shouldn't look to much on the graphs of integrands but rather think about their structure $\endgroup$ – tired Mar 9 '16 at 13:13
  • $\begingroup$ Still having a hard time integrating this. $\endgroup$ – Kenny Guy Mar 13 '16 at 9:59
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There exists no solution in terms of standard mathematical functions. This can be proven using Risch Algorithm. This is not something I recommend you do by hand.

The most easy way to check if an integral has a solution is by asking Wolfram Alpha which uses this algorithm or a better one to guarantee a solution is it exists. Otherwise, it displays:

(no result found in terms of standard mathematical functions)

To back up my claim:

For indefinite integrals, an extended version of the Risch algorithm is used whenever both the integrand and integral can be expressed in terms of elementary functions, exponential integral functions, polylogarithms, and other related functions.

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  • $\begingroup$ There is definitely a way to do this, we just have not found it yet. $\endgroup$ – Kenny Guy Mar 15 '16 at 4:40
  • $\begingroup$ @Kenny what makes you think that? $\endgroup$ – Jens Renders Mar 15 '16 at 7:03
  • $\begingroup$ Well there are yet a lot of things to discover and many concepts we do not know yet stays hidden. $\endgroup$ – Kenny Guy Mar 15 '16 at 7:08
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    $\begingroup$ What I'm trying to tell you is that we know for sure that the integral does not exist in terms of standard mathematical functions. Think of it like this, say we had no log, than 1/x would not have an integral. This can be proven using risch algorithm and wolfram alpha uses it. Of course you can give the answer using an infinite sum or definite integral, but thats easy to find for any integral and not what you are looking for I guess $\endgroup$ – Jens Renders Mar 15 '16 at 7:18
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I'm pretty sure this doesn't have a closed form. A change of variable and integration by parts got me to: $$\int \frac{dx}{1 + \arctan(x)} = \frac{x}{1 + \arctan(x)} + \int \frac{\tan(y)}{(1 + y)^2}dy$$ WolframAlpha says the right-most integral has no closed form, and I can't think of anything useful to do with it. It might have some cute definite integral values?

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  • $\begingroup$ Is there any way to integrate $\int \frac{\tan (y)}{(1+y)^2}dy$? $\endgroup$ – Kenny Guy Mar 14 '16 at 6:56
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    $\begingroup$ @KennyGuy most propably not! $\endgroup$ – tired Mar 14 '16 at 8:20

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