TITLE: Integral $\int_0^1 \frac{\arctan x}{x^2-x-1}dx$
QUESTION [8 upvotes]: 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.

REPLY [4 votes]: 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