Cfrgtkky

How can we find the value of $$int_0^1arctan xln(1+x)left(frac2x-frac3{1+x}right)dx$$ using elementary methods?

With some help of calculator I get the result: $displaystyle{frac3{128}pi^3-frac9{32}piln^22}$.

Thoughts of this integral

Since I have asked this question and Pisco gave a brilliant answer, I tried to convert $$I_1=int_0^1arctan xln(1+x)frac{dx}xtext{ and }I_2=int_0^1arctan xln(1+x)frac{dx}{1+x}$$ into the form of integral Pisco gave.

Integrating by parts to the second integral converts $I_2$ into $int_0^1frac{ln^2(1+x)}{1+x^2}dx$.

But for $I_1$? Integrating by parts gives a dilog function and I tried substitution $x=frac{1-t}{1+t}$ and got $$frac{lnfrac{2}{t+1} arctanfrac{1-t}{1+t}}{1-t^2}$$ which is not what I want.

Here is an elementary approach, although it turned into a crossover with FDP's answer.

First note that from here we have:
$$color{blue}{int_0^1 frac{arctan x ln(1+x)}{x}dx}=frac{3}{2}int_0^1 frac{arctan xln(1+x^2)}{x}dx$$
$$overset{IBP}=frac32 underbrace{ln xarctan xln(1+x^2)bigg|_0^1}_{=0}-frac32 left(int_0^1 frac{ln xln(1+x^2)}{1+x^2}dx+2int_0^1 frac{xarctan xln x}{1+x^2}dxright) $$
Back to the original integral, we have:
$$I=color{blue}{2int_0^1 frac{arctan x ln(1+x)}{x}dx}-color{red}{3int_0^1 frac{arctan x ln(1+x)}{1+x}dx} $$
$$=color{blue}{-3left(int_0^1 frac{ln xln(1+x^2)}{1+x^2}dx+2int_0^1 frac{xarctan xln x}{1+x^2}dxright)}-color{red}{3int_0^1frac{arctan xln(1+x)}{1+x}dx}$$
$$Rightarrow I=-3(B+2A+J)quad quad (1)$$
Where I kept the notation like in FDP's answer. Namely:
$$begin{align*}
displaystyle A&=int_0^1 dfrac{xarctan xln x}{1+x^2}dx\
displaystyle B&=int_0^1 dfrac{ln x ln(1+x^2)}{1+x^2}dx\
displaystyle J&=int_0^1dfrac{arctan xln(1+x)}{1+x}dx
end{align*}$$
Also another two identities follows from that post, see $(8)$ and $(9)$: $$J=dfrac{5}{3}Gln 2-dfrac{pi^3}{128}+dfrac{3pileft(ln 2right)^2}{32}+B+dfrac{2}{3}left(dfrac{Gln 2}{2}-dfrac{pi^3}{64}right)-dfrac{2}{3}cdotfrac{pi^3}{32} $$
$$Rightarrow color{purple}{J=2Gln 2 -frac{5pi^3}{128}+frac{3pi}{32}ln^2 2 +B} tag 2$$
$$color{magenta}{A=dfrac{1}{64}pi^3-B-Gln 2} tag 3$$
Now plugging $(2)$ and $(3)$ in $(1)$ yields:
$$I=-3left(B+2left(color{magenta}{dfrac{1}{64}pi^3-B-Gln 2}right)+ color{purple}{2Gln 2 -frac{5pi^3}{128}+frac{3pi}{32}ln^2 2 +B}right)$$
$$Rightarrow I=-3left(-frac{pi^3}{128}+frac{3pi}{32}ln^2 2right)=boxed{frac{3pi^3}{128}-frac{9pi}{32}ln^2 2}$$
Credits to FDP for his amazing answer there!

NOT A FULL SOLUTION BUT A START:

Here you have:

begin{equation}
I = int_0^1arctan xln(1+x)left(frac2x-frac3{1+x}right)dx = int_0^1arctan xln(1+x)left[frac{2 - x}{x(x + 1)}right]dx
end{equation}

Consider using Feynman's Trick with two parameters:

begin{equation}
I(a,b) = int_0^1arctan(ax)ln(1+bx)left[frac{2 - x}{x(x + 1)}right]dx
end{equation}

Here $I = I(1,1)$ and $I(0,b), I(a,0) = 0$. Here take the partial derivative with respect to $a$ and $b$ to yield:

begin{equation}
frac{partial^2I}{partial a partial b} = int_0^1frac{x}{a^2x^2 + 1}cdotfrac{x}{1 +bx}left[frac{2 - x}{x(x + 1)}right]dx = int_0^1 frac{xleft(2 - xright)}{left(a^2x^2 + 1right)left(1 + bxright)left(x + 1right)}dx
end{equation}

From here employ Partial Fraction Decomposition. I will finish off in an hour if you're still interested (sorry will be afk for the next hour).

For $I_1$, by integration by parts,
begin{eqnarray*}
I_1&=&int_0^1arctan xln(1+x)dln x\
&=&arctan xln(1+x)ln x|_0^1-int_0^1ln xleft(frac{ln(1+x)}{1+x^2}+frac{arctan x}{1+x}right)dx\
&=&-int_0^1frac{ln xln(1+x)}{1+x^2}dx-int_0^1frac{ln xarctan x}{1+x}dx\
&=&-I_3-I_4.
end{eqnarray*}
Here $I_3$ and $I_4$ are
$$ I_3=int_0^1frac{ln xln(1+x)}{1+x^2}dx, I_4=int_0^1frac{ln xarctan x}{1+x}dx. $$
From here,
$$ I_3= -2 G ln (2)-3 Imleft(text{Li}_3left(frac{1+i}{2}right)right)+frac{11 pi ^3}{128}+frac{3}{32} pi ln ^2(2). $$
From here,
$$ I_4=dfrac{Gln 2}{2}-dfrac{pi^3}{64} $$
$G$ is the Catalan's constant.