Integral

$\displaystyle \begin{aligned} I & = \int_0^1 \frac{1}{2^x + 1 + 2^{-x}} \ dx \\ & = \int_0^1 \frac{2^x}{2^{2x} + 2^x + 1} \ dx & \small \color{#3D99F6} \text{Let } u = 2^x \implies dx = \frac{1}{2^x \ln x} \ du \\ & = \frac{1}{\ln 2} \int_1^2 \frac{1}{u^2 + u + 1} \ dx \\ & = \frac{1}{\ln 2} \int_1^2 \frac{1}{\left(u + \frac 12\right)^2 + \frac 34} \ dx \\ & = \frac{1}{\ln 2} \cdot \frac{2}{\sqrt{3}} \tan ^{-1} \left( \frac{2u + 1}{\sqrt{3}} \right) \left. \right|_1^2 \\ & = \frac{2\pi}{3^{\frac 32} \ln 2} - \frac{2\tan ^{-1} \left(\frac{2}{\sqrt{3}}\right)}{\sqrt{3}\ln 2} \\ & \approx \boxed{0.3167} \end{aligned}$

$\large \displaystyle I = \int_{0}^{1} \frac{1}{1 + 2^{x} + 2^{-x}} dx$

$\large \displaystyle \color{#20A900} u = 2^x \rightarrow du = 2^x \ln(2) dx \rightarrow dx = \frac{1}{u\ln(2)}$

$\large \displaystyle I = \frac{1}{\ln(2)} \int_{1}^{2} \frac{1}{1 + u + u^{-1}} \frac{1}{u} du$

$\large \displaystyle I = \frac{1}{\ln(2)} \int_{1}^{2} \frac{1}{u^2 + u + 1} du$

$\large \displaystyle I = \frac{1}{\ln(2)} \int_{1}^{2} \frac{1}{(u+0.5)^2 + 0.75} du$

$\large \displaystyle \color{#20A900} v = u + 0.5 \rightarrow dv = du$

$\large \displaystyle I = \frac{1}{\ln(2)} \int_{1.5}^{2.5} \frac{1}{v^2 + 0.75} dv$

$\large \displaystyle I = \frac{1}{0.75 \ln(2)} \int_{1.5}^{2.5} \frac{1}{\left ( \frac{v}{\sqrt{0.75}} \right )^2 + 1} dv$

$\large \displaystyle I = \frac{1}{0.75 \ln(2)} \sqrt{0.75} \left [ \tan^{-1} \left (\frac{v}{\sqrt{0.75}} \right ) \right ] _{1.5}^{2.5}$

$\large \displaystyle I = \frac{1}{\sqrt{0.75} \ln(2)} \left [ \tan^{-1} \left ( \frac{5}{\sqrt{3}} \right ) - \tan^{-1} \left ( \frac{3}{\sqrt{3}} \right ) \right ]$

$\color{#3D99F6} \boxed{ \large \displaystyle I \approx 0.3167}$