anonymous
  • anonymous
Integrating two ways, one with a u sub only and one with a different u sub and then by parts I am out by more than just a constant. Integrand is x^5(x^3+1)^1/2 so first u sub was u=x^3 +1 and in second attempt u=x^3...help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
A somewhat tricky substitution: let \(u^{2/3}=x~\iff~u^2=x^3\), so that \(dx=\dfrac{2}{3}u^{-1/3}~du\). \[\int x^5\sqrt{x^3+1}~dx~~\iff~~\int u^{10/3}\sqrt{u^2+1}\left(\frac{2}{3}u^{-1/3}\right)~du\] Simplifying a bit leaves you with \[\frac{2}{3}\int u^{9/3}\sqrt{u^2+1}~du=\frac{2}{3}\int u^3\sqrt{u^2+1}~du\] which looks like it's easier to work with. You might have to find a similar but different substitution like the one above.
anonymous
  • anonymous
Or perhaps a trigonometric substitution will help right away: \(u=\tan t\), so you have \(du=\sec^2t~dt\) and \[\frac{2}{3}\int\tan^3t\sqrt{\tan^2t+1}~\sec^2t~dt=\frac{2}{3}\int\tan^3t\sec^3t~dt\] And this I'm sure can be done, with some effort.
anonymous
  • anonymous
I made a chrome extension that makes it easy to copy what other people have posted: \(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles A somewhat tricky substitution: let \(u^{2/3}=x~\iff~u^2=x^3\), so that \(dx=\dfrac{2}{3}u^{-1/3}~du\). \[\int x^5\sqrt{x^3+1}~dx~~\iff~~\int u^{10/3}\sqrt{u^2+1}\left(\frac{2}{3}u^{-1/3}\right)~du\] Simplifying a bit leaves you with \[\frac{2}{3}\int u^{9/3}\sqrt{u^2+1}~du=\frac{2}{3}\int u^3\sqrt{u^2+1}~du\] which looks like it's easier to work with. You might have to find a similar but different substitution like the one above. \(\color{blue}{\text{End of Quote}}\)

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anonymous
  • anonymous
@wio, neat! I'm surprised OS doesn't already have a feature like this
anonymous
  • anonymous
Yeah, it was something that I wish existed because for my more latex heavy posts it's hard to retype and correct them.
anonymous
  • anonymous
https://chrome.google.com/webstore/detail/open-study-quoting/daeffggacpioiaekojicffekajgcejgb
anonymous
  • anonymous
Sorry for invading @catanary. I hope you solve the problem.
anonymous
  • anonymous
I think we've got in under control. Thanks!
anonymous
  • anonymous
The substitution u=x^3 +1 then du = 3x^2 and x^3=u-1 gives 1/3∫(u-1)√u du which gives the correct answer \[\frac{ -2 }{ 9 }\left( x^{3} +1 \right)^{\frac{ 3 }{2 }}+\frac{ 2 }{ 15 }\left( x^{3} +1 \right)^{\frac{ 5 }{ 2 }}\] but I am not sure why when I let u = x^3 then du is 3x^2 so integral is 1/3∫u√u+1du and do this by parts I get an x^3 in my asnwer which is not supposed to be there.
anonymous
  • anonymous
Are you getting something like \[\frac{2}{3}x^3\left(x^3+1\right)^{3/2}-\frac{4}{15}\left(x^3+1\right)^{5/2}+C~~?\]
anonymous
  • anonymous
That' exactly what I am getting...Please can you tell me why its wrong?
goformit100
  • goformit100
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anonymous
  • anonymous
Hmm, according to Wolfram, the by-parts solution is off by only a factor of 3: http://www.wolframalpha.com/input/?i=D%5B%282%2F3%29*x%5E3*%28x%5E3%2B1%29%5E%283%2F2%29-%284%2F15%29%28x%5E3%2B1%29%5E%285%2F2%29%2Cx%5D I would say there's an algebraic error in your work and in mine, but I don't see where it is...

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