How to take $\int_1^{+\infty} \frac{dx}{x\sqrt{1+x^5+x^{10}}}$? I have the integral:
$$\int_1^{+\infty} \frac{dx}{x\sqrt{1+x^5+x^{10}}}$$
I have no clue how to approach it. i  was thinking of $x = \frac{1}{t}$ but it does not seem to be correct, or I do something wrong
 A: Sun $x=1/t$ and get
$$\int_0^1 dt \frac{t^4}{\sqrt{1+t^5+t^{10}}} = \frac15 \int_0^1 \frac{du}{\sqrt{1+u+u^2}}$$
You should be able to do the latter integral.  The result is $\frac15 \log{(1+2/\sqrt{3})}$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\int_{1}^{\infty}{\dd x \over x\root{1 + x^{5} + x^{10}}}
\,\,\,\stackrel{x^{5}\ \mapsto\ x}{=}\,\,\,
{1 \over 5}\int_{1}^{\infty}{\dd x \over x\root{1 + x + x^{2}}}
\\[5mm] & \stackrel{x\ =\ \pars{1 - t^{2}}/\pars{2t - 1}}{=}\,\,\,
{2 \over 5}\int_{\root{3} - 1}^{1/2}{\dd t \over t^{2} - 1} =
\bbx{{1 \over 5}\,\ln\pars{1 + {2\root{3} \over 3}}}\label{1}\tag{1}
\end{align}

The change of variables in \eqref{1} is Euler First Substitution.

