Integral $\int_0^1 \frac{\arctan x}{x^2-x-1}dx$ After seeing this integral I've decided to give a try to calculate:
$$I=\int_0^1 \frac{\arctan x}{x^2-x-1}dx$$
That is because it's common for many integrals to have a combination of a polynomial in the denominator and a logarithm or an inverse trig function in the numerator.
Mostly I tried standard ways such as integrating by parts, random substitutions, or using:
$$\frac{\arctan x}{x}=\int_0^1  \frac{dy}{1+x^2y^2}$$
But I realised that is not a great idea since it gives some mess after partial fractions, so I decided to prepare the integral a little for a Feynman's trick using probably the only helpful thing with that denominator, it won't change while using $x\mapsto 1-x$.
$$I=\int_0^1 \frac{\arctan x}{x^2-x-1}dx=\int_0^1 \frac{\arctan (1-x)}{x^2-x-1}dx$$
$$\Rightarrow 2I=\frac{\pi}{2 \sqrt 5}\ln\left(\frac{3-\sqrt 5}{3+\sqrt 5}\right)-\int_0^1 \frac{\arctan(x^2-x+1)}{x^2-x-1}dx$$
$$\small J(a)=\int_0^1 \frac{\arctan(a(x^2-x-1)+2)}{x^2-x-1}dx\Rightarrow J'(a)=\int_0^1 \frac{1}{1+(a(x^2-x-1)+2)^2}dx$$
But I am stuck now. I would like to see a method which finds a closed form for this integral, hopefully something decent comes out. I already imagine there will be some special functions.
 A: 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.
A: $$\small  \int_0^1 \frac{\arctan x}{x^2-x-1}dx\overset{x=\tan\frac{t}{2}}=-\frac12 \int_0^\frac{\pi}{2} \frac{t}{2+\tan t}\frac{dt}{\cos t}\overset{\tan t=  x}=-\frac12\int_0^\infty \frac{\arctan x}{(2+x)\sqrt{1+x^2}}dx$$
$$\small \overset{\large x=\frac{1/2+t}{1-t/2}}=-\frac{1}{2\sqrt 5}\int_{-\frac12}^2\frac{\arctan\left(\frac{\frac{1}{2}+t}{1-\frac{t}{2}}\right)}{\sqrt{1+t^2}}dt =-\frac{2}{\sqrt 5}\arctan\left(\frac12\right) \ln (\phi)-\frac{1}{2\sqrt 5}\int_\frac12^2 \frac{\arctan t}{\sqrt{1+t^2}}dt$$
$$\overset{\large t=\frac{x^2-1}{2x}}=-\frac{2}{\sqrt 5}\arctan\left(\frac12\right) \ln (\phi)+\frac{\pi}{2\sqrt 5}\ln(\phi)-\frac{1}{\sqrt 5}\int_{\phi}^{\phi^3}\frac{\arctan x}{x}dx$$
$$=\boxed{\frac{\ln(\phi)}{\sqrt 5}\arctan\left(\frac34\right)+\frac{1}{\sqrt 5}\left(\operatorname{Ti}_2\left(\phi\right)-\operatorname{Ti}_2\left(\phi^3\right)\right)}$$
$$\text{where} \ \phi=\frac{1+\sqrt 5}{2},\, \operatorname{Ti}_{2}(k)=\Im \operatorname{Li}_2(ik)=\int_0^k \frac{\arctan x}{x}dx.$$
A: A solution by Cornel I. Valean

We need a single main substitution $\displaystyle x\mapsto \frac{\varphi x-1/\varphi }{1+x}=g(x)$, which directly leads to 
  $$\mathcal{I}=-\frac{1}{\sqrt{5}}\int_{1/\varphi^2}^{\varphi^2}\frac{\arctan(g(x))}{x}\textrm{d}x=\frac{\arctan(1/\varphi)}{\sqrt{5}}\int_{1/\varphi^2}^{\varphi^2}\frac{1}{x}\textrm{d}x-\frac{1}{\sqrt{5}}\int_{1/\varphi}^{\varphi^3}\frac{\arctan(x)}{x}\textrm{d}x,$$
  where the last integral can be expressed using the inverse tangent integral, $\displaystyle \operatorname{Ti}_2(x)=\int_0^x\frac{\arctan(t)}{t}\textrm{d}t$.

End of story
A: Here is my attempt. However, it does not result in a typical "nice, closed form."
Let $I$ be the original integral. Let $x^2-x-1 = 0$. Solving for $x$ using the quadratic formula gives us $x \in \left\{\frac{1+\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2}\right\}$, which means we can rewrite $x^2-x-1$ as $\left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right)$.
Using partial fraction decomposition on $\frac{1}{\left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right)}$, Let $A$ and $B$ be some arbitrary constants such that
$$
\frac{1}{\left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right)} = \frac{A}{\left(x-\frac{1+\sqrt{5}}{2}\right)} + \frac{B}{\left(x-\frac{1-\sqrt{5}}{2}\right)}.
$$
Multiplying both sides by the denominator of the left side gives us
$$
1 = A\left(x-\frac{1-\sqrt{5}}{2}\right) + B\left(x-\frac{1+\sqrt{5}}{2}\right).
$$
We solve for $A$ and $B$. If $x = \frac{1+\sqrt{5}}{2}$, then solving for $A$ yields
\begin{align*}
    1 &= A\left(\frac{1+\sqrt{5}}{2}-\frac{1-\sqrt{5}}{2}\right) + B\left(\frac{1+\sqrt{5}}{2}-\frac{1+\sqrt{5}}{2}\right) \\
    &= A\left(\frac{1+\sqrt{5}}{2}-\frac{1-\sqrt{5}}{2}\right) \\
    &= A\sqrt{5} \\
    \frac{1}{\sqrt{5}} &= A. \\
\end{align*}
Similarly, if $x = \frac{1-\sqrt{5}}{2}$, then solving for $B$ yields
\begin{align*}
    1 &= A\left(\frac{1-\sqrt{5}}{2}-\frac{1-\sqrt{5}}{2}\right) + B\left(\frac{1-\sqrt{5}}{2}-\frac{1+\sqrt{5}}{2}\right) \\
    &= B\left(\frac{1-\sqrt{5}}{2}-\frac{1+\sqrt{5}}{2}\right) \\
    &= B\sqrt{5} \\
    \frac{1}{\sqrt{5}} &= B. \\
\end{align*}
Therefore, our partial fraction decomposition results in
$$
\frac{1}{\left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right)} = \frac{1}{\sqrt{5}}\left(\frac{1}{x-\frac{1+\sqrt{5}}{2}}\right) + \frac{1}{\sqrt{5}}\left(\frac{1}{x-\frac{1-\sqrt{5}}{2}}\right)
$$
Thus, our integral becomes
$$
I = \frac{1}{\sqrt{5}}\int_{0}^{1}\frac{\arctan\left(x\right)}{x-\frac{1+\sqrt{5}}{2}}d\theta-\frac{1}{\sqrt{5}}\int_{0}^{1}\frac{\arctan\left(x\right)}{x-\frac{1-\sqrt{5}}{2}}d\theta.
$$
Next, let's define these two integrals and two constants:
$$
I_1 = \int_{0}^{1}\frac{\arctan\left(x\right)}{x-\phi}d\theta \text{ and }
I_2 = \int_{0}^{1}\frac{\arctan\left(x\right)}{x-\theta}d\theta,
$$
such that
$$
\phi = \frac{1+\sqrt{5}}{2} \text{ and } \theta = \frac{1-\sqrt{5}}{2}.
$$
Both $I_1$ and $I_2$ are special functions. Namely, they are Generalized Inverse Tangent Functions $\text{Ti}_2\left(1, -\phi\right)$ and $\text{Ti}_2\left(1, -\theta\right)$, respectively (https://mathworld.wolfram.com/InverseTangentIntegral.html).
Therefore, our integral $I$ is
$$
\int_0^{1}\frac{\arctan{x}}{x^2-x-1}d\theta = \frac{1}{\sqrt{5}}\left(\text{Ti}_2\left(1, -\frac{1+\sqrt{5}}{2}\right) - \text{Ti}_2\left(1,  \frac{\sqrt{5}-1}{2}\right)\right).
$$
