Krdytkyu

The following integral was proposed by Cornel Ioan Valean and appeared as Problem $12054$ in the American Mathematical Monthly earlier this year.

Prove $$int_0^1 frac{arctan x}{x}lnleft(frac{1+x^2}{(1-x)^2}right)dx=frac{pi^3}{16}$$

I had small tries for it, such as:
Letting $x=tan t$ which gives:
$$I=int_0^1 frac{arctan x}{x}lnleft(frac{1+x^2}{(1-x)^2}right)dx=-int_0^frac{pi}{4}frac{t}{sin t}ln(1-sin(2t))dt=$$
$$overset{2t=x}=-frac12 underset{=J}{int_0^frac{pi}{2}frac{x}{sin x} ln(1-sin x)dx}=-frac12 int_0^frac{pi}{2} xln(1-sin x) left(lnleft(tan frac{x}{2}right)right)'dx=$$
$$overset{IBP}=frac12int_0^frac{pi}{2} lnleft(1-sin xright)lnleft(tan frac{x}{2}right)dx+frac12 int_0^frac{pi}{2} frac{xcos x}{sin x-1}lnleft(tan frac{x}{2}right)dx$$
Or to employ Feynman's trick for the first integral $(J)$ in the second row. $$J(t)=int_0^frac{pi}{2} frac{xln(1-tsin x)}{sin x}dxRightarrow J'(t)=int_0^frac{pi}{2} frac{x}{1-tsin x}dx$$ But even so I don't see a how to obtain a closed from for the last one. Also with a different parameter: $$J(t)=int_0^frac{pi}{2} frac{text{arccot} (t cot x)ln(1-sin x)}{sin x}dx$$
$$Rightarrow J'(t)=-int_0^frac{pi}{2} frac{ln(1-sin x)cos x}{1+t^2 cot^2x}frac{dx}{sin^2x}overset{sin x=y}=int_0^1 frac{ln(1-y)}{1+t^2left(1-frac{1}{y^2}right)}frac{dy}{y^2}$$

Also from here we have the following relation:
$$int_0^1 frac{arctan x ln(1+x^2)}{x} dx =frac23 int_0^1 frac{arctan x ln(1+x)}{x}dx$$
Thus we can rewrite the integral as:
$$I=frac23 int_0^1 frac{arctan x ln(1+x)}{x}dx -2int_0^1 frac{arctan x ln(1-x)}{x}dx$$
$$=frac23 int_0^1 int_0^1 frac{ln(1+x)-3ln(1-x)}{1+x^2y^2}dydx=frac23 int_0^1 int_t^1 frac{ln(1+x)-3ln(1-x)}{1+t^2}dxdt $$

Another option might be to rewrite:
$$lnleft(frac{1+x^2}{(1-x)^2}right)= lnleft(frac{1+x}{1-x}right)+lnleft(frac{1+x^2}{1-x^2}right)$$
$$Rightarrow I= int_0^1 frac{arctan x}{x}lnleft(frac{1+x}{1-x}right)dx+int_0^1 frac{arctan x}{x}lnleft(frac{1+x^2}{1-x^2}right)dx$$
And now to use the power expansion of the log functions inside to obtain:
$$I=sum_{n=0}^infty frac{2}{2n+1}int_0^1 frac{arctan x}{x} , left(x^{2n+1}+x^{4n+2}right)dx=sum_{n=0}^infty frac{2}{2n+1}int_0^1int_0^1 frac{left(x^{2n+1}+x^{4n+2}right)}{1+y^2x^2}dydx$$

In the meantime I found one nice solution by Roberto Tauraso here.

This seems like an awesome integral and I would like to learn more so I am searching for more approaches.
Would any of you who also already solve it and submitted the answer to the AMM or know how to solve this integral kindly share the solution here?

Edit: Another impressive solution due to Yaghoub Sharifi is found here.

Another approach,

Perform integration by parts,

begin{align*}
I&=int_0^1 frac{arctan x}{x}lnleft(frac{1+x^2}{(1-x)^2}right),dx\
&=Big[ln (x) lnleft(frac{1+x^2}{(1-x)^2}right)arctan xBig]_0^1 -int_0^1 frac{ln x}{1+x^2}lnleft(frac{1+x^2}{(1-x)^2}right)dx-int_0^1 frac{2(1+x)ln (x)arctan (x)}{(1-x)(1+x^2)}dx\
&=-int_0^1 frac{ln x}{1+x^2}lnleft(frac{1+x^2}{(1-x)^2}right)dx-2int_0^1 frac{(1+x)ln (x)arctan (x)}{(1-x)(1+x^2)}dx\
end{align*}

For $xin [0;1]$ define the function $R$ by,

begin{align*}
R(x)=int_0^x frac{(1+t)ln t}{(1-t)(1+t^2)}dt=int_0^1 frac{x(1+tx)ln (tx)}{(1-tx)(1+t^2x^2)}dt\
end{align*}

Observe that,

begin{align*}
R(1)=int_0^1 frac{tln t}{1+t}dt+int_0^1 frac{ln t}{1-t}dt
end{align*}
Perform integration by parts,

begin{align*}
I&=-int_0^1 frac{ln x}{1+x^2}lnleft(frac{1+x^2}{(1-x)^2}right)dx-2Big[R(x)arctan xBig]_0^1+2int_0^1frac{R(x)}{1+x^2}dx\
&=-int_0^1 frac{ln x}{1+x^2}lnleft(frac{1+x^2}{(1-x)^2}right)dx-frac{pi}{2}R(1)+2int_0^1 int_0^1 frac{x(1+tx)ln (tx)}{(1-tx)(1+t^2x^2)(1+x^2)}dtdx\
&=-int_0^1 frac{ln x}{1+x^2}lnleft(frac{1+x^2}{(1-x)^2}right)dx-frac{pi}{2}R(1)+int_0^1 ln xleft[frac{1}{1+x^2}lnleft(frac{1+t^2x^2}{(1-tx)^2}right)right]_{t=0}^{t=1} dx+\
&int_0^1 ln tleft[frac{1}{1+t^2}lnleft(frac{1+x^2}{(1-tx)^2}right)+frac{2arctan (tx)}{1-t^2}-frac{2tarctan x}{1+t^2}-frac{2tarctan x}{1-t^2}right]_{x=0}^{x=1} dt\
&=-frac{pi }{2}R(1)+ln 2int_0^1 frac{ln t}{1+t^2}dt-2int_0^1 frac{ln (1-t)ln t}{1+t^2}dt+2int_0^1 frac{ln tarctan t}{1-t^2}dt-\
&frac{pi}{2} int_0^1 frac{tln t}{1+t^2}dt-frac{pi}{2} int_0^1frac{tln t}{1-t^2} dt\
end{align*}

For $xin [0;1]$ define the function $S$ by,

begin{align*}
S(x)=int_0^x frac{ln t}{1-t^2}dt=int_0^1 frac{xln(tx)}{1-t^2x^2} dt
end{align*}

Perform integration by parts,

begin{align*}
int_0^1 frac{ln xarctan x}{1-x^2}dx&=Big[S(x)arctan xBig]_0^1-int_0^1 frac{S(x)}{1+x^2}dx\
&=frac{pi}{4}S(1)-int_0^1 int_0^1 frac{xln(tx)} {(1-t^2x^2)(1+x^2)} dtdx\
&=frac{pi}{4}S(1)-frac{1}{2}int_0^1 left[ frac{ln x}{1+x^2}lnleft(frac{1+tx}{1-tx} right)right]_{t=0}^{t=1} dx-\
&frac{1}{2}int_0^1 left[ frac{ln t}{1+t^2}lnleft(frac{1+x^2}{1-t^2x^2} right)right]_{x=0}^{x=1}dt\
&=frac{pi}{4}S(1)-frac{ln 2}{2}int_0^1 frac{ln t}{1+t^2}dt+int_0^1 frac{ln(1-x)ln x}{1+x^2}dx
end{align*}

Therefore,

begin{align*}I&=piint_0^1frac{2tln t}{t^4-1} dtend{align*}

Perform the change of variable $y=t^2$,

begin{align*}I&=frac{1}{2}pi int_0^1 frac{ln y}{y^2-1}dy\
&=frac{1}{2}pitimes frac{3}{4}zeta(2)\
&=frac{pi^3}{16}
end{align*}

Put
begin{equation*}
I=int_{0}^1dfrac{arctan x}{x}lnleft(dfrac{1+x^2}{(1-x)^2}right), mathrm{d}x.
end{equation*}
Via the substitution $ x=dfrac{z}{z+1}$ we get
begin{equation*}
I = int_{0}^{infty}dfrac{arctan frac{z}{z+1}ln(2z^2+2z+1)}{z^2+z}, mathrm{d}z.
end{equation*}
Put
begin{equation*}
log z=ln|z|+iarg z, quad -pi<arg z <pi.
end{equation*}
Then
begin{equation*}
arctan frac{z}{z+1}ln(2z^2+2z+1) = text{Im}left(log^2(1+z+iz)right).
end{equation*}
Consequently
begin{equation*}
I = text{Im}left(int_{0}^{infty}dfrac{log^2(1+z+iz)}{z^2+z}right)mathrm{d}z.
end{equation*}
However, $ log(z) $ is an analytic function in $ text{Re} z>0 $. According to Cauchys integral theorem we get the same value if we integrate along the curve with the parametrization $ z=(1-i)s, s>0 $.
begin{gather*}
I = text{Im}left(int_{0}^{infty}dfrac{ln^2(2s+1)}{s(s+1-is)}, mathrm{d}sright) = int_{0}^{infty}dfrac{ln^2(2s+1)}{2s^2+2s+1}, mathrm{d}s = \[2ex] int_{0}^{infty}dfrac{2ln^2(2s+1)}{(2s+1)^2+1}, mathrm{d}s = [t=2s+1] = \[2ex] int_{1}^{infty}dfrac{ln^2(t)}{t^2+1}, mathrm{d}t =[u= 1/t] = int_{0}^{1}dfrac{ln^2(u)}{u^2+1}, mathrm{d}u.
end{gather*}
Thus
begin{equation*}
2I = int_{0}^{infty}dfrac{ln^2(u)}{u^2+1}, mathrm{d}u
end{equation*}
In order to evaluate this integral we integrate $ dfrac{log^3(z)}{z^2+1} $ along a keyhole contour and use residue calculus.
In this case $ log z =ln |z|+iarg z, quad 0<arg z < 2pi $.
We get
begin{equation*}
I = dfrac{pi^3}{16}.
end{equation*}