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Do integration by parts to get $$I=\underbrace{\int_0^1\frac{\tan^{-1}(x)}{x}\ln(1+x)\, dx}_{J}-{\underbrace{\int_0^1\frac{
\ln(x)}{1+x}\tan^{-1}(x)\, dx}_{K}}$$
The first integral is solved in this Instagram post. For $K$,
$$K=\int_0^1\frac{\tan^{-1} x}{x}\ln(x)-{\underbrace{\int_0^1\int_0^1\frac{\ln x}{(1+x)(1+x^2y^2)}\, dx\, dy}_{K'}}$$
$$K'=\int_0^1\int_0^y\frac{\ln(\frac{x}{y})}{(x+y)(1+x^2)}\, dxdy$$
$$=\int_0^1\int_0^y\frac{\ln(\frac{x}{y})}{(x+y)(1+x^2)}+\frac{\ln(\frac{y}{x})}{(x+y)(1+y^2)}\, dxdy-\int_0^1 \int_0^1 \frac{\ln(\frac{y}{x})}{(x+y)(1+y^2)}\, dx\, dy$$
$$K'=\frac{1}{2}\int_0^1\int_0^1 \left(\frac{\ln(\frac{x}{y})}{(x+y)(1+x^2)}+\frac{\ln(\frac{y}{x})}{(x+y)(1+y^2)}\right)dx\, dy+\int_0^1\frac{\ln x}{1+x}\, dx\int_0^1 \frac{dy}{1+y^2}$$
$$=\frac{1}{2}\int_0^1\int_0^1 \frac{y\ln(x)+x\ln(y)-y\ln(y)-x\ln(x)}{(1+y^2)(1+x^2)}\, dx\, dy+\frac{\pi}{4}\frac{\pi^2}{12}$$
Now the integrals can be evaluated individually, we get
$$K=\frac{\ln 2}{2}G-\frac{\pi^3}{64}$$
P.S. Since not everyone has an Instagram account, I might as well put my solution here to make the answer more self contained.
Consider $$I_1=\int_0^1\frac{\tan^{-1} x}{x}\ln(1-x)\, dx \quad,\quad I_2=\int_0^1\frac{\tan^{-1} x}{x}\ln(1+x)\, dx$$
Denote $\chi(x)=\int_0^x \frac{\tanh^{-1} z}{z}\, dz$.
Note the result $$\chi (z)+\chi\left(\frac{1-z}{1+z}\right)=\frac{\pi^2}{8}-\frac{1}{2}\ln(z)\ln\left(\frac{1-z}{1+z}\right)$$
Which could be seen true by differentiating both sides.
\par
Multiply both sides by $\frac{1}{1+z^2}$ and integrate both sides form 0 to 1
$$2\int_0^1 \frac{\chi(x)}{1+x^2}\, dx=\frac{\pi^3}{32}+\int_0^1 \frac{\ln x}{1+x^2}\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\, dx$$
The left hand side, via IBP is equal to:
$$2\left[\frac{\pi^2}{8}\frac{\pi}{4}-\int_0^1\frac{\tan^{-1} x\tanh^{-1} x}{x}\, dx\right]$$
$$=\frac{\pi^3}{16}-I_2+I_1$$
i.e. $$I_1-I_2=-\frac{\pi^3}{32}+{\underbrace{\int_0^1 \frac{\ln x}{1+x^2}\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\, dx}_{I'}}$$
We compute $I'$:
$$I'=\int_0^1 \frac{\ln x}{1+x^2}\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\, dx$$
$$=\int_0^1 \int_0^1 \frac{z \ln z}{(1+z^2)(1-z^2 t^2)}\, dz\, dt=\int_0^1 \frac{1}{1+t^2}\int_0^1 \left(\frac{t^2 z \ln z}{1-z^2 t^2}+\frac{z \ln z}{1+z^2}\right)\, dz\, dt$$
$$=\int_0^1 \frac{1}{1+t^2}{\underbrace{\int_0^1 \frac{t^2 z \ln z}{1-z^2 t^2}\, dz}_{-\frac{\text{Li}_2(t^2)}{4}}}\, dt+\int_0^1 \frac{z\ln z}{1+z^2}\, dz \int_0^1 \frac{dt}{1+t^2}$$
$$=-\frac{\pi^3}{192}-\frac{1}{4}\int_0^1 \frac{\dilogarithm{t^2}}{1+t^2}\, dt$$
doing IBP gives
$$=-\frac{\pi^3}{192}-\frac{1}{4}\left(\zeta(2)\frac{\pi}{4}+\int_0^1 2\frac{\ln(1-t^2)}{t}\tan^{-1}(t)\, dt\right)$$
$$\implies I'=-\frac{\pi^3}{64}-\frac{I_1}{2}-\frac{I_2}{2}$$
Putting this back gives
$$I_1-I_2=-\frac{\pi^3}{32}+I'$$
$$I_1-I_2=-\frac{3\pi^3}{64}-\frac{I_1}{2}-\frac{I_2}{2}$$
$$\implies I_2=\frac{3\pi^3}{32}+3I_1$$Hence we have the relationship between $I_2$ and $I_1$
We now compute $I_1$:
$$I_1=\int_0^1 \frac{\tan^{-1} x}{x}\ln(1-x)\, dx$$
$$=\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\int_0^1 x^{2n}\ln(1-x)\, dx=-{\underbrace{\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}H_{2n+1}}_{S}}$$
To compute $S$, consider the following: By differentiating both sides, we can see that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2}x^n=\text{Li}_3({x})-\text{Li}_3({1-x})+\ln(1-x)\text{Li}_2({1-x})+\frac{1}{2}\ln^2(1-x)\ln x+\zeta(3)$$ Then by setting $x=i$ and taking the imaginary part, we get alternating odd sums, Which is equal to $S$.
$$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}H_{2n+1}=-\Im \text{Li}_3(1-i)-\frac{\pi}{16}\ln^2(2)-\frac{G}{2}\ln(2)$$
Where $\Im$ denotes 'the imaginary part of'. Note that $\Im \text{Li}_3(1-i)\approx -1.26708\cdots$
\begin{equation}
I_1=\frac{\pi}{16}\ln^2(2)+\frac{G}{2}\ln(2)+\Im(\trilogarithm{1-i})
\end{equation}
which also gives the result of $I_2$
\begin{equation}
I_2=\frac{3\pi^3}{32}+\frac{3\pi}{16}\ln^2(2)+\frac{3G}{2}\ln(2)+3\Im(\trilogarithm{1-i})
\end{equation}
