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
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1$\begingroup$ Have you tried $x^5=t$? then $5dx=x^{-4} dt$ and you'll get something like $\frac{dt}{t\sqrt{1+t+t^2}}$. $\endgroup$ – Mikhail Tikhonov May 21 '17 at 12:02
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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 $\frac15 \log{(1+2/\sqrt{3})}$.
answered May 21 '17 at 12:05
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$\begingroup$ I don't know what I am doing wrong since I get $\frac{1}{5} \log \left(1+\frac{2}{\sqrt{3}}\right)$ $\endgroup$ – Claude Leibovici May 21 '17 at 14:01
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$\begingroup$ @ClaudeLeibovici: you did nothing wrong - I did. Messed up the integration limits. $\endgroup$ – Ron Gordon May 21 '17 at 14:33
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$\begingroup$ This was just a very minor detail ! The solution you provided is really nice. Cheers. $\endgroup$ – Claude Leibovici May 21 '17 at 14:45
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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.
answered May 21 '17 at 22:08
