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My attempt:

$$\int \arctan^2x \,\mathrm dx=\arctan x\left(x \arctan x-\dfrac{\ln|1+x^2|}{2}\right)-\int\dfrac{\left(x \arctan x-\frac{\ln|1+x^2|}{2}\right)}{1+x^2}\,\mathrm dx$$

I tried calculating this first

$$\displaystyle\int\dfrac{x\arctan x}{1+x^2}dx$$

For this last integral, let $u=\arctan x, \mathrm dx=\mathrm du(1+x^2)$ , then

$$\int\dfrac{x\arctan x}{1+x^2}\,\mathrm dx=\displaystyle\int u\tan u\,\mathrm du$$

For $\displaystyle\int u\tan u\,\mathrm du$ first try:

$$\displaystyle\int u\tan u\,\mathrm du=-u\ln|\cos u|+\displaystyle\int \ln|\cos u|\,\mathrm du$$

I couldn't do anything with this integral (only tried $\tan(u/2)=j$).

For $\displaystyle\int u\tan u \,\mathrm du$ second try:

$$\displaystyle\int u\tan u \,\mathrm du=\frac{u^2}{2}\tan u-\dfrac12\displaystyle\int u^2\sec^2u \,\mathrm du$$

Then we clearly see, we have nothing.

And... I couldn't even calculate $\dfrac12\displaystyle\int\dfrac{\ln|1+x^2|}{1+x^2}\,\mathrm dx$

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    $\begingroup$ It is not an elementary integral, it is related with $\text{Li}_2(x)$ and the Clausen function: en.wikipedia.org/wiki/Clausen_function $\endgroup$ – Jack D'Aurizio Sep 29 '16 at 21:38
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    $\begingroup$ However, many definite integrals over peculiar intervals have nice closed forms due to some reflection formulas. For instance, $$\int_{0}^{1}\arctan^2(x)\,dx = -K+\frac{\pi^2}{16}+\frac{\pi\log 2}{4}.$$ $\endgroup$ – Jack D'Aurizio Sep 29 '16 at 21:41
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Write the arc tangent using logarithms to get: $$I:= \int \arctan^2x \,\mathrm d x=-\frac 14\int \left( \ln (1+xi)-\ln(1-xi)\right)^2\,\mathrm d x=\\ \int \ln^2(1+xi)\,\mathrm dx-2\int \ln (1-xi)\ln(1+xi)\,\mathrm d x+\int \ln^2(1-xi)\,\mathrm d x $$


The first integral can be solved by substituting $u=1+xi$ and then integrating by parts: $$I_1:=\int \ln^2(1+xi)\,\mathrm dx=-i\int\ln^2(u)=-i\left(2\int \ln u\,\mathrm d u+u\ln^2u\right)=-i\left( u\ln^2u-2u\ln u+2u\right)=\boxed{-iu\ln^2u+2iu\ln u-2iu}$$


The third integral is a bit more difficult but similar - substitute $u=-1+xi$, $v=\ln u$ and then integrate by parts twice to get the result: $$I_3:=\ln^2(1-xi)\,\mathrm d x=-i\int(\ln u+\ln(-1))^2\,\mathrm d u=-i\int e^v(v+\pi i)^2 \,\mathrm d v=\\ 2i\int e^v(v+\pi i)\,\mathrm d v-ie^v(v+\pi i)=\boxed{-ie^v\left((v+\pi i)^2-2(v+\pi i)+2\right)}$$


The second integral however is much trickier and messier. First, integrate by parts: $$-2I_2:=-2\int\ln(1-xi)\ln(1+xi)\,\mathrm d x \\ I_2=\int\ln(1-xi)\ln(1+xi)\,\mathrm d x=-\int\frac{(x-i)\ln(1+xi)-x+i}{x+i}\,\mathrm d x-\\ i(1+xi)\ln(1-xi)\left( \ln(1+xi)-1\right) $$

$$-J:=-\int\frac{(x-i)\ln(1+xi)-x+i}{x+i}\,\mathrm d x$$

Expand: $$J=\int\frac{(x-i)\ln(1+xi)}{x+i}\,\mathrm d x-\int\frac{x}{x+i}\,\mathrm dx+i\int\frac{1}{x+i}\,\mathrm dx=\int\frac{(x-i)\ln(1+xi)}{x+i}\,\mathrm d x+\\ (1-i)(\ln(x+i))+x$$

Subsitute $u=x+i$ and expand again: $$K:= \int\frac{(x-i)\ln(1+xi)}{x+i}\,\mathrm d x =\int\frac{(u-2i)\left( \ln(u-2i)+\frac\pi 2 i\right)}{u}\,\mathrm d u=\\ -2i\int\frac{\ln(u-2i)}{u}\,\mathrm d u+\int\ln(u-2i)\,\mathrm d u-\pi\ln u+\frac\pi 2 iu $$

The second integral is solved immediately after substituting $v=u-2i$: $$K_2:=\int\ln(u-2i)\,\mathrm d u=v\ln v-v=(u-2i)\ln(u-2i)-u+2i$$

To solve the first one we can rewrite the integrand: $$-2iK_1:=-2i\int\frac{\ln(u-2i)}{u}\,\mathrm d u \\ K_1=\int \frac{\ln\left( \frac{i}{2}u+1\right)}{u}\,\mathrm d u+\left( \log 2-\frac{\pi}{2}i\right)(\ln u)$$

and now substitute $v=-\frac i2 u$ to finish by seeing that the integral is a dilogarithm: $$\int \frac{\ln\left( \frac{i}{2}u+1\right)}{u}\,\mathrm d u=-\int -\frac{\ln(1-v)}{v}\,\mathrm d v=-\mathrm{Li}_2 (v)=-\mathrm{Li}_2\left( -\frac{i}{2}u\right)$$


And thus our integral is solved. Plug in all the previously solved integrals and undo the substitutions to get the final, messy result which I won't be writing out here.

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Let $x\in {\mathbb R}$ then integrating by parts we have: \begin{eqnarray} \int \arctan(x)^2 dx =x \arctan(x)^2 - \log(1+x^2) \arctan(x) + \frac{1}{(2 \imath) }\sum\limits_{\xi\in\{-1,1\}} \xi \left( \log(\imath \xi+ x) \log(\frac{1}{2}(1+\imath \xi x))+ Li_2(\frac{1}{2}(1-\imath \xi x)) - \frac{1}{2} \log(x+ \imath \xi)^2\right) \quad (i) \end{eqnarray}

In[1023]:= (*Integrate[Log[1+x^2]/(1+x^2],x]*)
f[x_] := 1/(2 I) Sum[
    xi (Log[I xi + x] Log[1/2 (1 + I xi x)] + 
       PolyLog[2, 1/2 (1 - I xi x)] - 1/2 Log[x + I xi]^2), {xi, -1, 
     1, 2}];
D[x ArcTan[x]^2 - Log[1 + x^2] ArcTan[x] + f[x], x] // Simplify

Out[1024]= ((1 + x^2) ArcTan[x]^2 + Log[-I + x] + Log[I + x] - 
 Log[1 + x^2])/(1 + x^2)

Therefore for example: \begin{equation} \int\limits_{-1}^1 \arctan(x)^2 dx = \frac{1}{8} (\pi (\pi +\log (16))-16 C) \end{equation} where $C$ is the Catalan constant.

Now having said all this if $x$ is complex things get much more complicated since, firstly $\log(x+\imath) + \log(x-\imath) \neq \log(1+x^2)$ and secondly the quantitities in the right hand side in $(i)$ can be discontinuous and therefore we need to compute any definite integrals by avoiding discontinuities, i.e. as a principal value.

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