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

With some help of calculator I get the result: $\displaystyle{\frac3{128}\pi^3-\frac9{32}\pi\ln^22}$.
Thoughts of this integral
Since I have asked this question and Pisco gave a brilliant answer, I tried to convert $$I_1=\int_0^1\arctan x\ln(1+x)\frac{dx}x\text{ and }I_2=\int_0^1\arctan x\ln(1+x)\frac{dx}{1+x}$$ into the form of integral Pisco gave.
Integrating by parts to the second integral converts $I_2$ into $\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx$.
But for $I_1$? Integrating by parts gives a dilog function and I tried substitution $x=\frac{1-t}{1+t}$ and got $$\frac{\ln\frac{2}{t+1} \arctan\frac{1-t}{1+t}}{1-t^2}$$ which is not what I want.
 A: Here is an elementary approach, although it turned into a crossover with FDP's answer.
First note that from here we have:
$$\color{blue}{\int_0^1 \frac{\arctan x \ln(1+x)}{x}dx}=\frac{3}{2}\int_0^1 \frac{\arctan x\ln(1+x^2)}{x}dx$$
$$\overset{IBP}=\frac32 \underbrace{\ln x\arctan x\ln(1+x^2)\bigg|_0^1}_{=0}-\frac32 \left(\int_0^1 \frac{\ln x\ln(1+x^2)}{1+x^2}dx+2\int_0^1 \frac{x\arctan x\ln x}{1+x^2}dx\right) $$
Back to the original integral, we have:
$$I=\color{blue}{2\int_0^1 \frac{\arctan x \ln(1+x)}{x}dx}-\color{red}{3\int_0^1 \frac{\arctan x \ln(1+x)}{1+x}dx} $$
$$=\color{blue}{-3\left(\int_0^1 \frac{\ln x\ln(1+x^2)}{1+x^2}dx+2\int_0^1 \frac{x\arctan x\ln x}{1+x^2}dx\right)}-\color{red}{3\int_0^1\frac{\arctan x\ln(1+x)}{1+x}dx}$$
$$\Rightarrow I=-3(B+2A+J)\quad \quad  (1)$$
Where I kept the notation like in FDP's answer. Namely:
$$\begin{align*}
\displaystyle A&=\int_0^1 \dfrac{x\arctan x\ln x}{1+x^2}dx\\
\displaystyle B&=\int_0^1 \dfrac{\ln x \ln(1+x^2)}{1+x^2}dx\\
\displaystyle J&=\int_0^1\dfrac{\arctan x\ln(1+x)}{1+x}dx
\end{align*}$$
Also another two identities follows from that post, see $(8)$ and  $(9)$: $$J=\dfrac{5}{3}G\ln 2-\dfrac{\pi^3}{128}+\dfrac{3\pi\left(\ln 2\right)^2}{32}+B+\dfrac{2}{3}\left(\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64}\right)-\dfrac{2}{3}\cdot\frac{\pi^3}{32} $$
$$\Rightarrow \color{purple}{J=2G\ln 2 -\frac{5\pi^3}{128}+\frac{3\pi}{32}\ln^2 2 +B} \tag 2$$
$$\color{magenta}{A=\dfrac{1}{64}\pi^3-B-G\ln 2} \tag 3$$
Now plugging $(2)$ and $(3)$ in $(1)$ yields:
$$I=-3\left(B+2\left(\color{magenta}{\dfrac{1}{64}\pi^3-B-G\ln 2}\right)+ \color{purple}{2G\ln 2 -\frac{5\pi^3}{128}+\frac{3\pi}{32}\ln^2 2 +B}\right)$$
$$\Rightarrow I=-3\left(-\frac{\pi^3}{128}+\frac{3\pi}{32}\ln^2 2\right)=\boxed{\frac{3\pi^3}{128}-\frac{9\pi}{32}\ln^2 2}$$
Credits to FDP for his amazing answer there!
A: NOT A FULL SOLUTION BUT A START:
Here you have:
\begin{equation}
 I = \int_0^1\arctan x\ln(1+x)\left(\frac2x-\frac3{1+x}\right)dx = \int_0^1\arctan x\ln(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_0^1\arctan(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_0^1\frac{x}{a^2x^2 + 1}\cdot\frac{x}{1 +bx}\left[\frac{2 - x}{x(x + 1)}\right]dx = \int_0^1 \frac{x\left(2 - x\right)}{\left(a^2x^2 + 1\right)\left(1 + bx\right)\left(x + 1\right)}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). 
A: For $I_1$, by integration by parts,
\begin{eqnarray*}
I_1&=&\int_0^1\arctan x\ln(1+x)d\ln x\\
&=&\arctan x\ln(1+x)\ln x|_0^1-\int_0^1\ln x\left(\frac{\ln(1+x)}{1+x^2}+\frac{\arctan x}{1+x}\right)dx\\
&=&-\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}dx-\int_0^1\frac{\ln x\arctan x}{1+x}dx\\
&=&-I_3-I_4.
\end{eqnarray*}
Here $I_3$ and $I_4$ are
$$ I_3=\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}dx, I_4=\int_0^1\frac{\ln x\arctan x}{1+x}dx. $$
From here,
$$ I_3= -2 G \ln (2)-3 \Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)+\frac{11 \pi ^3}{128}+\frac{3}{32} \pi  \ln ^2(2). $$
From here,
$$ I_4=\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64} $$
$G$ is the Catalan's constant. 
A: $$I=2\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx-3\int_0^1\frac{\arctan x\ln(1+x)}{1+x}\ dx$$
Apply IBP for the second integral, we get, 
$$I=2\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx-\frac{3\pi}{8}\ln^22+\frac32\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx$$
The first integral was calculated here :

$$\int_0^1 \frac{\arctan x\ln(1+x)}{x}\ dx=\frac{3\pi^3}{32}+\frac{3\pi}{16}\ln^22+\frac32G\ln2-3\Im\operatorname{Li}_3(1+i)$$

And the second integral was calculated here:

$$\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx=4\Im\operatorname{Li}_3(1+i)-\frac{7\pi^3}{64}-\frac{3\pi}{16}\ln^22-2G\ln2$$

Substituting these two results, we get $\quad\displaystyle \boxed{I={\frac{3\pi^3}{128}-\frac{9\pi}{32}\ln^2 2}}$
