Dtjytyk

My attempt:

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

I tried calculating this first

$$displaystyleintdfrac{xarctan x}{1+x^2}dx$$

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

$$intdfrac{xarctan x}{1+x^2},mathrm dx=displaystyleint utan u,mathrm du$$

For $displaystyleint utan u,mathrm du$ first try:

$$displaystyleint utan u,mathrm du=-uln|cos u|+displaystyleint ln|cos u|,mathrm du$$

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

For $displaystyleint utan u ,mathrm du$ second try:

$$displaystyleint utan u ,mathrm du=frac{u^2}{2}tan u-dfrac12displaystyleint u^2sec^2u ,mathrm du$$

Then we clearly see, we have nothing.

And... I couldn't even calculate $dfrac12displaystyleintdfrac{ln|1+x^2|}{1+x^2},mathrm dx$

Write the arc tangent using logarithms to get:
$$I:= int arctan^2x ,mathrm d x=-frac 14int left( ln (1+xi)-ln(1-xi)right)^2,mathrm d x=\
int ln^2(1+xi),mathrm dx-2int ln (1-xi)ln(1+xi),mathrm d x+int ln^2(1-xi),mathrm d x
$$

The first integral can be solved by substituting $u=1+xi$ and then integrating by parts:
$$I_1:=int ln^2(1+xi),mathrm dx=-iintln^2(u)=-ileft(2int ln u,mathrm d u+uln^2uright)=-ileft( uln^2u-2uln u+2uright)=boxed{-iuln^2u+2iuln u-2iu}$$

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:
$$I_3:=ln^2(1-xi),mathrm d x=-iint(ln u+ln(-1))^2,mathrm d u=-iint e^v(v+pi i)^2 ,mathrm d v=\
2iint e^v(v+pi i),mathrm d v-ie^v(v+pi i)=boxed{-ie^vleft((v+pi i)^2-2(v+pi i)+2right)}$$

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

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

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

Subsitute $u=x+i$ and expand again:
$$K:= intfrac{(x-i)ln(1+xi)}{x+i},mathrm d x =intfrac{(u-2i)left( ln(u-2i)+fracpi 2 iright)}{u},mathrm d u=\
-2iintfrac{ln(u-2i)}{u},mathrm d u+intln(u-2i),mathrm d u-piln u+fracpi 2 iu
$$

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

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

and now substitute $v=-frac i2 u$ to finish by seeing that the integral is a dilogarithm:
$$int frac{lnleft( frac{i}{2}u+1right)}{u},mathrm d u=-int -frac{ln(1-v)}{v},mathrm d v=-mathrm{Li}_2 (v)=-mathrm{Li}_2left( -frac{i}{2}uright)$$

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.

Let $xin {mathbb R}$ 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) }sumlimits_{xiin{-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)^2right) 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}
intlimits_{-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.