Some Calculus!

My solution is similar, but not identical.

We have $\int_0^1 \frac{\ln x}{x+a}\,dx \; = \; -\tfrac16\pi^2 - \tfrac12\ln^2a + \tfrac12\ln^2(1+a) + \mathrm{Li}_2\left(\tfrac{a}{1+a}\right) \hspace{1cm} a > 0$ and hence, with $\phi = \tfrac12(\sqrt{5}+1)$ being the Golden Section, $\begin{aligned} \int_0^1 \frac{\ln x}{x+\phi}\,dx & = -\tfrac16\pi^2 + \tfrac32\ln^2\phi + \mathrm{Li}_2(\phi^{-1}) \\ \int_0^1 \frac{\ln x}{x + \phi^{-1}}\,dx & = -\tfrac16\pi^2 + \mathrm{Li}_2(\phi^{-2}) \end{aligned}$ and hence $\begin{aligned} \int_0^1 \frac{\ln x}{x^2 + \sqrt{5}x + 1}\,dx & = \int_0^1 \frac{\ln x}{(x+\phi)(x+\phi^{-1})}\,dx \; = \; \mathrm{Li}_2(\phi^{-2}) - \mathrm{Li}_2(\phi^{-1}) - \tfrac32\ln^2\phi \\ & = \left(\tfrac{1}{15}\pi^2 - \tfrac14\ln^2(\phi^{-2})\right) - \left(\tfrac{1}{10}\pi^2 - \ln^2(\phi^{-1})\right) - \tfrac32\ln^2\phi \\ & = -\tfrac{1}{30}\pi^2 - \tfrac32\ln^2\phi \end{aligned}$ using standard properties of the dilogarithm function $\mathrm{Li}_2$ .

Now consider the cut plane with the negative real axis removed, and the key-hole contour $\Gamma = \gamma_1 - \gamma_2 - \gamma_3 + \gamma_4$ , where

• $\gamma_1$ is the (almost) circular arc $z = e^{ix}$ for $-\pi < x < \pi$ ,
• $\gamma_2$ is the straight-line segment $z = x e^{\pi i}$ for $\epsilon < x < 1$ , just above the cut,
• $\gamma_3$ is the (almost) circular arc $z = \epsilon e^{ix}$ for $-\pi < x < \pi$ ,
• $\gamma_4$ is the straight-line segment $z = x e^{-\pi i}$ for $\epsilon < x < 1$ , just below the cut.

The function $f(z) \; = \; \frac{(\log z)^2}{z^2 - \sqrt{5}z + 1}$ is analytic inside and on the curve $\Gamma$ , except for a simple pole at $z = \phi^{-1}$ . Now $\begin{aligned} \int_{\gamma_1} f(z)\,dz & = \int_{-\pi}^\pi \frac{(ix)^2 i e^{ix}\,dx}{e^{ix}(2\cos x - \sqrt{5})} \; = \; 2i\int_0^1 \frac{x^2\,dx}{\sqrt{5} - 2\cos x} \\ \left(\int_{\gamma_2} - \int_{\gamma_4}\right)f(z)\,dz & = -\int_0^1 \frac{(\ln x + \pi i)^2 - (\ln x - \pi i)^2}{x^2 + \sqrt{5}x + 1}\,dx \\ & = -4\pi i\int_0^1 \frac{\ln x}{x^2 + \sqrt{5}x + 1}\,dx \\ \int_{\gamma_3}f(z)\,dz & = O\big(\epsilon \ln\epsilon) \hspace{1cm} \epsilon \to 0 \end{aligned}$ Thus, since $\left(\int_{\gamma_1} - \int_{\gamma_2} - \int_{\gamma_3} + \int_{\gamma_4}\right)f(z)\,dz \; = \; 2\pi i \mathrm{Res}_{z=\phi^{_1}} f(z) \; = \; -2\pi i \ln^2(\phi^{-1}) \; = \; -2\pi i \ln^2\phi$ we deduce, letting $\epsilon \to 0$ , that $\int_0^\pi \frac{x^2\,dx}{\sqrt{5} - 2\cos x} + 2\pi\int_0^1\frac{\ln x}{x^2 + \sqrt{5}x + 1}\,dx \; = \; -\pi\ln^2(\phi^{-1}) \; = \; -\pi\ln^2\phi$ and hence $\int_0^\pi \frac{x^2\,dx}{\sqrt{5} - 2\cos x} \; = \; \tfrac{1}{15}\pi^3 + 2\pi\ln^2\phi$ which makes the answer $3 + 15 + 2 - 1 + 5 - 2 = \boxed{22}$ .

Obviously this is not an original question.(Wish i had that kind of brains)

Credit goes to china(tian27546).

Solution