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?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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With
  $\ds{t \equiv \root{x^{2} + x + 1} - x \implies
x = -\,{t^{2} - 1 \over 2t - 1}}$:

\begin{align}
\int_{1}^{\infty}{\dd x \over x\root{x^{2} + x + 1}} & =
-2\int_{\root{3} - 1}^{1/2}\,\,\,{\dd t \over 1 - t^{2}} =
\int_{\root{3} - 1}^{1/2}\pars{-\,{1 \over 1 - t} - {1 \over 1 + t}}\,\dd t
\\[5mm] & =
\left.\ln\pars{1 - t \over 1 + t}\right\vert_{\ \root{3}\ -\ 1}^{\ 1/2} =
\ln\pars{{1 - 1/2 \over 1 + 1/2}\,
{\bracks{\root{3} - 1} + 1 \over 1 - \bracks{\root{3} - 1}}}
\\[5mm] & =
\bbx{\ds{\ln\pars{1 + {2 \over 3}\,\root{3}}}} \approx 0.7677
\end{align}
A: We have:
$$ \int\frac{dx}{(x+1)\sqrt{x^2+3x+3}}=C+\log(x+1)-\log\left(3+x+2\sqrt{x^2+3x+3}\right)\tag{1}$$
hence:
$$ \int_{1}^{+\infty}\frac{dx}{x\sqrt{x^2+x+1}} = \color{blue}{\log\left(1+\frac{2}{\sqrt{3}}\right)}.\tag{2}$$
To check this, it is enough to notice that
$$ I=\int_{1}^{+\infty}\frac{dx}{x\sqrt{x^2+x+1}}=2\int_{3}^{+\infty}\frac{dx}{(x-1)\sqrt{x^2+3}}\tag{3}$$
and by setting $x=\sqrt{3}\sinh z$ in the last integral,
$$ I = 2\int_{\text{arcsinh}\sqrt{3}}^{+\infty}\frac{dz}{\sqrt{3}\sinh z-1}\tag{4} $$
that is converted into the integral of a simple rational function by the substitution $z=\log u$.
A: $$
\begin{aligned}
\int_1^\infty \frac{dx}{x\sqrt{x^2+x+1}} & \stackrel{x\mapsto\frac{1}{x}}{=}\int_0^1 \frac{d x}{\sqrt{1+x+x^2}} \\
&=\int_0^1 \frac{d x}{\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}}
\end{aligned}
$$
Letting $x+\frac{1}{2} =\frac {\sqrt 3}{ 2 }\sinh \theta$ transforms the integral into
$$
\begin{aligned}
\int_1^\infty \frac{dx}{x\sqrt{x^2+x+1}} &=\int_{\sinh ^{-1}\left(\frac{1}{\sqrt{3}}\right)}^{\sinh ^{-1}(\sqrt{3})} \frac{\frac{\sqrt{3}}{2} \cosh \theta}{\frac{\sqrt{3}}{2} \cosh \theta} d \theta \\
&=\sinh ^{-1}(\sqrt{3})-\sinh ^{-1}\left(\frac{1}{\sqrt{3}}\right) \\
& \approx 0.76765
\end{aligned}
$$
