Calculate $\int_0^1\arctan x\left(\frac{3\ln(1+x^2)}{1+x}-\frac{2\ln(1+x)}{x}\right)\ dx$ Prove that 

$$\int_0^1\arctan x\left(\frac{3\ln(1+x^2)}{1+x}-\frac{2\ln(1+x)}{x}\right)\ dx=\frac{3\pi}{8}\ln^22-\frac{\pi^3}{32}$$

I managed to prove the above equality using integral manipulation (solution be posted soon) , but is it possible to do it in different ways and specifically by harmonic series?
The interesting thing about this problem is that we don't see any imaginary part which is usually involved in such integrals. 
Note: the second integral $$\int_0^1 \frac{\arctan x\ln(1+x)}{x}\ dx=\frac{3\pi^3}{32}+\frac{3\pi}{16}\ln^22+\frac32G\ln2+3\text{Im}\operatorname{Li}_3(1-i)$$ was already evaluated here but I calculated the original problem without separating the two integrals. 
 A: @Kemono Chen elegantly proved here
$$\int_0^y\frac{\ln(1+yx)}{1+x^2}dx=\frac12 \tan^{-1}(y)\ln(1+y^2)$$
By integration by parts we have
$$\int_0^y\frac{y\tan^{-1}(x)}{1+yx}dx=\frac12 \tan^{-1}(y)\ln(1+y^2)$$
divide both sides by $1+y$ and integrate from $y=0$ to $y=1$, we get,
\begin{align}
\frac12I&=\frac12\int_0^1\frac{\arctan y\ln(1+y^2)}{1+y}\ dy=\int_0^1\int_0^y\frac{y\arctan x}{(1-yx)(1+y)}\ dy\ dx\\
&=\int_0^1\arctan x\left(\int_x^1\frac{y}{(1-yx)(1+y)}\ dy\right)\ dx\\
&=\int_0^1\arctan x\left(\frac{\ln(1+x)}{x}-\frac{\ln(1+x^2)}{x}+\frac{2\ln(1+x)-\ln(1+x^2)-\ln2}{1-x}\right)\ dx\\
&=\underbrace{\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx}_{J}-\underbrace{\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx}_{K}\\
&\quad+\underbrace{\int_0^1\frac{\arctan x}{1-x}\left(2\ln(1+x)-\ln(1+x^2)-\ln2\right)\ dx}_{\large{x=(1-y)/(1+y)}}\\
&=J-K-\int_0^1\frac{\left(\frac{\pi}{4}-\arctan x\right)}{x(1+x)}\ln(1+x^2)\ dx\\
&=J-K-\frac{\pi}{4}\int_0^1\frac{\ln(1+x^2)}{x(1+x)}\ dx+K-I\\
\frac32I&=J-\frac{\pi}{4}\left(\frac{3\pi^2}{48}-\frac34\ln^22\right)\\
3I-2J&=\frac{3\pi}{8}\ln^22-\frac{\pi^3}{32}
\end{align}
