Value of $\int_1^\infty \frac{dx}{x\sqrt{x^2+x+1}}$?

Question

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?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}

Answer 2

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 $z=\log u$.

Answer 3

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