anonymous
  • anonymous
integral sqrt(x+1)/x^2 dx from 1 to infinite to compute and prove that it's convergent. for the convergence i know that if lim from x to inf from the value of the integral is a real number the integral converges. the computation is the issue here.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
So you need to figure out whether the following integral converges: \[\int_1^\infty \frac{\sqrt{x+1}}{x^2}~dx~~?\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
A u-sub should work nicely here. Let \(u=\sqrt{x+1}\), so that you get \(du=\dfrac{1}{2\sqrt{x+1}}dx\), or equivalently, \(dx=2u~du\). From the initial substitution, you also know that \(u^2=x+1\), which tells you \(x=u^2-1\). Keep track of your limits: upper: \(x\to\infty~~\Rightarrow~~u\to\infty\). lower: \(x=1~~\Rightarrow~~u=\sqrt{1+1}=\sqrt2\). So you have \[\large \int_\sqrt2^\infty \frac{u}{\left(1-u^2\right)^2}\bigg(2u~du\bigg)\\ \large 2\int_\sqrt2^\infty \frac{u^2}{\left(1-u^2\right)^2}~du\\ \]

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anonymous
  • anonymous
Partial fractions from here.
anonymous
  • anonymous
You can simplify this somewhat with another substitution. It'd make the PFD easier on the eyes, but not necessary.
anonymous
  • anonymous
as in replace the u^2 w/ say v and go w/ partial fractions from there?
anonymous
  • anonymous
Something like that, yeah. It'd involve a square root, though. I'd stick with this first sub for now though.
anonymous
  • anonymous
btw, when you did the substitution you gave the integrand \[\frac{ u }{ (1-u^2)^2 }\]. wasn't it supposed to be \[\frac{ u }{ (u^2 - 1)^2 }\]?
anonymous
  • anonymous
did i miss something there?
anonymous
  • anonymous
Maybe some color coding will help. Just a sec.
anonymous
  • anonymous
Sorry, had some connection troubles.
anonymous
  • anonymous
Got it now. Re-did the entire integral myself now and figured out where it all fits. Ty very much.
anonymous
  • anonymous
\[\int_1^\infty\frac{\sqrt{x+1}}{x^2}~dx\] \(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles A u-sub should work nicely here. Let \(\color{red}{u=\sqrt{x+1}}\), so that you get \(\color{blue}{du=\dfrac{1}{2\sqrt{x+1}}dx}\), or equivalently, \(\color{blue}{dx=2u~du}\). From the initial substitution, you also know that \(\color{purple}{u^2=x+1}\), which tells you \(\color{purple}{x=u^2-1}\). Keep track of your limits: upper: \(x\to\infty~~\Rightarrow~~u\to\infty\). lower: \(x=1~~\Rightarrow~~u=\sqrt{1+1}=\sqrt2\). So you have \[\large \int_\sqrt2^\infty \frac{u}{\left(1-u^2\right)^2}\bigg(2u~du\bigg)\\ \large 2\int_\sqrt2^\infty \frac{u^2}{\left(1-u^2\right)^2}~du\\ \] \(\color{blue}{\text{End of Quote}}\) \[\int_1^\infty \frac{\color{red}{\sqrt{x+1}}}{\color{purple}{x}^2}~\color{blue}{dx} =2\int_\sqrt2^\infty \frac{\color{red}{u}}{\left(\color{purple}{u^2-1}\right)^2}~\color{blue}{2u~du}=2\int_\sqrt2^\infty \frac{u^2}{\left(u^2-1\right)^2}~du\]
anonymous
  • anonymous
You're welcome

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