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.

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
$$
\int_0^{1}\frac{\arctan{x}}{x^2-x-1}d\theta = \frac{1}{\sqrt{5}}\left(\text{Ti}_2\left(1, -\frac{1+\sqrt{5}}{2}\right) - \text{Ti}_2\left(1,  \frac{\sqrt{5}-1}{2}\right)\right).
$$
