Integrate $\int \frac{1}{1+\arctan(x)}dx$ 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.
 A: 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.

A: 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?
