Mad exponent and integration

If $\phi = \tfrac12(1 + \sqrt{5})$ is the Golden Ratio, then $\phi^2 = \phi + 1$ , so that $3+\sqrt{5} = 2\phi^2$ , and so the integral is $I \; = \; \int_0^1 \frac{2^x}{(2\phi)^x + (2\phi^2)^{x}} \; = \; \int_0^1 \frac{dx}{\phi^x + \phi^{2x}} \; = \; \int_0^1 \frac{\phi^x}{\phi^{2x} + \phi^{3x}}\,dx$ and hence the substitution $u = \phi^x$ yields $\begin{aligned} I & = \; \frac{1}{\ln\phi}\int_1^\phi \frac{du}{u^2(1+u)} \; = \; \frac{1}{\ln\phi}\int_1^\phi \left(\frac{1}{u^2} - \frac{1}{u} + \frac{1}{u+1}\right)\,du \\ & = \; \frac{1}{\ln\phi}\Big[-u^{-1} + \ln\left(\frac{u+1}{u}\right)\Big]_1^\phi \; = \; \frac{1}{\ln\phi}\Big[\frac{\phi-1}{\phi} + \ln(\tfrac12\phi\big)\Big] \\ & = \; \boxed{0.353338...} \end{aligned}$

$\large y = \left(\frac{1+\sqrt{5}}{2}\right)^{x} , \frac{dy}{dx} = \left(\frac{1+\sqrt{5}}{2}\right)^x\ln\left(\frac{1+\sqrt{5}}{2}\right)$

$\large \int_{0}^{1}\frac{2^{x}}{(1+\sqrt{5})^{x}+(3+\sqrt{5})^{x}} dx$

$\large = \int_{0}^{1}\frac{2^{x}}{(2y)^{x}+(2y^2)^{x}} dx$

$\large = \left(\ln\left(\frac{1+\sqrt{5}}{2}\right)\right)^{-1} \int_{1}^{\frac{1+\sqrt{5}}{2}}\frac{1}{y^2(y+1)}dy$

$\large = \left(\ln\left(\frac{1+\sqrt{5}}{2}\right)\right)^{-1} \int_{1}^{\frac{1+\sqrt{5}}{2}}\frac{1}{y^2}+\frac{1}{y+1}-\frac{1}{y} dy$

$\large = \left(\ln\left(\frac{1+\sqrt{5}}{2}\right)\right)^{-1}\left[-\frac{1}{y}+\ln(y+1)-\ln(y)\right]_{1}^{\frac{1+\sqrt{5}}{2}}$

$\large = 0.353 \text{ (corr. to 3 sig. fig. ) }$