Value of $\int_1^\infty \frac{dx}{x\sqrt{x^2+x+1}}$? Not so thrilling... An exercise of one of my daughters.
How to evaluate 
$$\int_1^\infty \frac{dx}{x\sqrt{x^2+x+1}}?$$ I made several substitution namely:


*

*Factorisation of $x^2+x+1$

*Then use of $\sinh t$

*Then substitution by $e^u$

*To get a rational fraction with at the denominator a degree two polynomial with two real roots that can be integrated with partial fraction decomposition.


Is there something more straight forward?

$$
\begin{aligned}
\int_1^\infty \frac{dx}{x\sqrt{x^2+x+1}} & \stackrel{x\mapsto\frac{1}{x}}{=}\int_0^1 \frac{d x}{\sqrt{1+x+x^2}} \\
&=\int_0^1 \frac{d x}{\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}}
\end{aligned}
$$
Letting $x+\frac{1}{2} =\frac {\sqrt 3}{ 2 }\sinh \theta$ transforms the integral into
$$
\begin{aligned}
\int_1^\infty \frac{dx}{x\sqrt{x^2+x+1}} &=\int_{\sinh ^{-1}\left(\frac{1}{\sqrt{3}}\right)}^{\sinh ^{-1}(\sqrt{3})} \frac{\frac{\sqrt{3}}{2} \cosh \theta}{\frac{\sqrt{3}}{2} \cosh \theta} d \theta \\
&=\sinh ^{-1}(\sqrt{3})-\sinh ^{-1}\left(\frac{1}{\sqrt{3}}\right) \\
& \approx 0.76765
\end{aligned}
$$
