1
$\begingroup$

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

$\endgroup$
1
  • 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
6
$\begingroup$

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 $\frac15 \log{(1+2/\sqrt{3})}$.

$\endgroup$
4
  • 1
    $\begingroup$ Excellent $(+1)$ $\endgroup$ – tired May 21 '17 at 12:06
  • $\begingroup$ 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
  • $\begingroup$ @ClaudeLeibovici: you did nothing wrong - I did. Messed up the integration limits. $\endgroup$ – Ron Gordon May 21 '17 at 14:33
  • $\begingroup$ This was just a very minor detail ! The solution you provided is really nice. Cheers. $\endgroup$ – Claude Leibovici May 21 '17 at 14:45
1
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.