Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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.
\(\color{blue}{\text{End of Quote}}\)

Answer 4

@wio, neat! I'm surprised OS doesn't already have a feature like this

Answer 5

Yeah, it was something that I wish existed because for my more latex heavy posts it's hard to retype and correct them.

Answer 6

https://chrome.google.com/webstore/detail/open-study-quoting/daeffggacpioiaekojicffekajgcejgb

Answer 7

Sorry for invading @catanary. I hope you solve the problem.

Answer 8

I think we've got in under control. Thanks!

Answer 9

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.

Answer 10

Are you getting something like
\[\frac{2}{3}x^3\left(x^3+1\right)^{3/2}-\frac{4}{15}\left(x^3+1\right)^{5/2}+C~~?\]

Answer 11

That' exactly what I am getting...Please can you tell me why its wrong?

Answer 12

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Answer 13

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...