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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?

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  • $\begingroup$ use the function $$arctanh$$ $\endgroup$ – Dr. Sonnhard Graubner Dec 17 '16 at 19:55
  • $\begingroup$ $x^2+x+1$ can't be factorised using real numbers - only complex numbers, which in turn makes the integral harder. Please update your question to show all your working - not just the descriptions. $\endgroup$ – unseen_rider Dec 17 '16 at 19:57
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    $\begingroup$ Would $$x^2+x+1=\left(x+\frac12\right)^2+\frac34$$ help? $\endgroup$ – J. M. is a poor mathematician Dec 17 '16 at 20:07
  • $\begingroup$ @J.M. That's what I mean in the first step I mentioned. I realize that my wording is probably incorrect as this is not a proper factorization.I don't know what is the proper English wording that describes the transformation $a x^2 +b x +c$ into $ a(x+\frac{b}{2a})^2+c -\left(\frac{b}{2a}\right)^2$ $\endgroup$ – mathcounterexamples.net Dec 17 '16 at 20:11
  • $\begingroup$ Yes, it's often called "completing the square" in English. From there, you can use either the tangent or the hyperbolic sine as seen fit. $\endgroup$ – J. M. is a poor mathematician Dec 17 '16 at 20:13
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}

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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 $z=\log u$.

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    $\begingroup$ I don't see what is getting easier here. How do you perform the integration? I mean, the original function has a primitive $\log x-\log(2+x+2\sqrt{1+x+x^2})$, but if we just write that, then we do not show how it was obtained... $\endgroup$ – mickep Dec 17 '16 at 20:00
  • $\begingroup$ @amWhy: is it better now? $\endgroup$ – Jack D'Aurizio Dec 17 '16 at 20:18
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    $\begingroup$ @JackD'Aurizio This is almost what I did, but then you have the simple rational fraction that is not so simple... I know simple is a relative question! $\endgroup$ – mathcounterexamples.net Dec 17 '16 at 20:21
  • $\begingroup$ @JackD'Aurizio On my side I get for the denominator $\sqrt {3} u^2 -\sqrt{3}-u$??? $\endgroup$ – mathcounterexamples.net Dec 17 '16 at 20:34
  • $\begingroup$ @mathcounterexamples.net: all right, but it makes little difference. Factor that second-degree polynomial in order to have a partial fraction decomposition of the integrand function, leading to some logarithmic contributes. $\endgroup$ – Jack D'Aurizio Dec 17 '16 at 20:36

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