Here is one approach. As a warning, the final answer I find is not pretty.
Let
$$I = \int_0^1 \frac{\tan^{-1} x}{x^2 -x - 1} \, dx.$$
Start by using a self-similar substitution of
$$x = \frac{1 - u}{1 + u}, \,\, dx = -\frac{2}{(1 + u)^2} \, du.$$
So , after having reverted the dummy variable $u$ back to $x$, we have
$$I = 2 \int_0^1 \frac{\tan^{-1} \left (\frac{1 - x}{1 + x} \right )}{x^2 - 4x - 1} \, dx.$$
Noting that for $0 < x < 1$
$$\tan^{-1} \left (\frac{1 - x}{1 + x} \right ) = \frac{\pi}{4} - \tan^{-1} x,$$
then
\begin{align}
I &= \frac{\pi}{2} \int_0^1 \frac{dx}{x^2 - 4x - 1} - 2 \int_0^1 \frac{\tan^{-1} x}{x^2 - 4x - 1} \, dx\\
&= -\frac{\pi}{2 \sqrt{5}} \coth^{-1} \left (\frac{3}{\sqrt{5}} \right ) - 2 J,
\end{align}
where
$$J = \int_0^1 \frac{\tan^{-1} x}{x^2 - 4x - 1} \, dx.$$
To find $J$ we begin by noting that $\tan^{-1} x = \operatorname{Im} \ln (1 + ix)$. Thus
$$J = \operatorname{Im} \int_0^1 \frac{\ln (1 + ix)}{x^2 - 4x - 1} \, dx.$$
Making a substitution of $t = 1 + ix$ we have
$$J = - \operatorname{Re} \int_1^{1+i} \frac{\ln t}{(t - \alpha)(t - \beta)} \, dt,$$
where $\alpha = 1 + i(2 - \sqrt{5})$ and $\beta = 1 + i(2 + \sqrt{5})$. After performing a partial fraction decomposition we are left with
$$J = \frac{1}{2 \sqrt{5}} \operatorname{Im} \left [\int_1^{1 + i} \frac{\ln t}{\alpha - t} \, dt - \int_1^{1 + i} \frac{\ln t}{\beta - t} \, dt \right ].$$
Now, as
$$\int \frac{\ln x}{z - x} \, dx = - \ln \left (1 - \frac{x}{z} \right ) \ln x - \operatorname{Li}_2 \left (\frac{x}{z} \right ),$$
(for a proof of this result see the appendix below), one has
$$J = \frac{1}{2 \sqrt{5}} \operatorname{Im} \left [\ln (1 + i) \ln \left (\frac{\alpha}{\beta} \cdot \frac{\beta - 1 - i}{\alpha - i - i} \right ) + \operatorname{Li}_2 \left (\frac{1}{\alpha} \right ) - \operatorname{Li}_2 \left (\frac{1}{\beta} \right ) + \operatorname{Li}_2 \left (\frac{1 + i}{\beta} \right ) - \operatorname{Li}_2 \left (\frac{1 + i}{\alpha} \right ) \right ]$$
or after performing a huge amount of algebra
\begin{align}
J &= \frac{\pi}{8 \sqrt{5}} \ln (\sqrt{5} - 1) + \frac{1}{2 \sqrt{5}} \operatorname{Im} \left [\operatorname{Li}_2 \left (\frac{1}{2} + \frac{1}{\sqrt{5}}+ \frac{i}{2 \sqrt{5}} \right ) - \operatorname{Li}_2 \left (\frac{1}{2} - \frac{1}{\sqrt{5}} - \frac{i}{2 \sqrt{5}} \right ) \right.\\
& \quad+ \left. \operatorname{Li}_2 \left (\frac{1}{2} -\frac{1}{2 \sqrt{5}} - i \left (\frac{3}{2 \sqrt{5}} - \frac{1}{2} \right ) \right ) - \operatorname{Li}_2 \left (\frac{1}{2} +\frac{1}{2 \sqrt{5}} + i \left (\frac{3}{2 \sqrt{5}} + \frac{1}{2} \right ) \right ) \right ]\\
&= \frac{\pi}{8 \sqrt{5}} \ln (\sqrt{5} - 1) + \frac{1}{2 \sqrt{5}} \operatorname{Im} \frak{w},
\end{align}
where $\frak{w}$ is the term containing the four dilogarithms with complex arguments. Thus
$$\int_0^1 \frac{\tan^{-1} x}{x^2 - x - 1} \, dx = -\frac{\pi}{4 \sqrt{5}} \left (\ln 2 + \sinh^{-1} (2) \right ) - \frac{1}{\sqrt{5}} \operatorname{Im} \frak{w}.$$
Note that as $\operatorname{Im} {\frak{w}} = -0.8363170651979\ldots$ we see that $I \approx -0.376513$.
Appendix
Proof of
$$\int \frac{\ln x}{z - x} \, dx = - \ln \left (1 - \frac{x}{z} \right ) \ln x - \operatorname{Li}_2 \left (\frac{x}{z} \right ) + C$$
Setting $t = x/z, dt = dx/z$, we have
\begin{align}
\int \frac{\ln x}{z - x} \, dx &= \int \frac{\ln (zt)}{1 - t} \, dt\\
&= -\ln (1 - t) \ln (zt) + \int \frac{\ln (1 - t)}{t} \, dt \qquad \text{(by parts)}\\
&= -\ln (1 - t) \ln (zt) - \operatorname{Li}_2 (t) + C\\
&= - \ln \left (1 - \frac{x}{z} \right ) \ln x - \operatorname{Li}_2 \left (\frac{x}{z} \right ) + C,
\end{align}
as required to show.
When you evaluate an improper integral where its bound is a singularity, you don't take the limit as $x$ approaches that bound on both sides, as you can see from the definition of improper integrals. In our case, the integral $\displaystyle \int_{0}^{1}e^{x}\ln\left(x\right)dx$ is improperly integrable because $\displaystyle \lim_{a\to 0^+} \int_{a}^{1}e^{x}\ln\left(x\right)dx$ converges.
For this quick calculation, we will use the Puiseux Series of the Exponential Integral
$$\operatorname{Ei}(x) = \gamma + \ln(x) + \sum_{n=1}^{\infty}\frac{x^{n}}{nn!}.$$
Indeed,
$$
\begin{align}
\int_{0}^{1}e^{x}\ln\left(x\right)dx &= \lim_{a\to 0^+} \int_{a}^{1}e^{x}\ln\left(x\right)dx \\
&= \lim_{a\to 0^+}\Big[e^x \ln(x) - \operatorname{Ei}(x)\Big]_{a}^{1} \\
&= e^1 \ln(1) - \operatorname{Ei}(1) - \lim_{a\to 0^+} \left(e^a \ln(a) - \operatorname{Ei}(a)\right) \\
&= -\operatorname{Ei}(1) - \lim_{a\to 0^+}\left(e^a \ln(a) - \left(\gamma + \ln(a) + \sum_{n=1}^{\infty}\frac{a^n}{nn!}\right)\right) \\
&= -\operatorname{Ei}(1) - \lim_{a\to 0^+} \left(e^a \ln(a) - \ln(a)\right) + \lim_{a\to 0^+} \sum_{n=1}^{\infty}\frac{a^n}{nn!} + \lim_{a\to 0^+}\gamma \\
&= -\operatorname{Ei}(1) - 0 + 0 + \gamma \\
&= \gamma - \operatorname{Ei}(1). \\
\end{align}
$$
Best Answer
Integrate by parts
\begin{align} \int_0^1 \ln x\operatorname{Li}_2(x)\ dx &= \int_0^1 \operatorname{Li}_2(x)\ d[x(\ln x-1)]\\ &=-\text{Li}_2(1) +\int_0^1 (\ln x-1)\ln(1-x)dx\\ &=-\frac{\pi^2}6 +\left(3-\frac{\pi^2}6 \right)=3-\frac{\pi^2}3 \end{align}