How do you even write a problem so you can show how far you can get on your own?

J

joeyjon

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Apr 13, 2011
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how do you integrate (the entire equation is the square root of)


((x^(3)-3)/x^(11))^(1/2)
 
Well, I figure everything under the radical is the u for the substitution, so the derivative of that is

33x^(-12)-8x^(-9)dx

but that doesn't seem to help, I am not sure what to do now.

I think I need to come up with a different u. So do I just take shots in the dark now?

maybe try x^(3)-3 as my u?
 
Oh and what really confuses me is that the back of the book has the answer (2/27)(1-(3/x^(3))^(3/2)

so does that mean they used 1-(3/x^(3)) as there u somehow? How did they ever decide or pick that as the u?
 
wait I am onto something (x^(-8)(1-3x^(-3))^(1/2)

huh. I guess I'll keep trying to guess my way into the right answer. I wish my book or the teacher would have explained this.

So now I can use that as the u...
 
\(\displaystyle \displaystyle \int \sqrt{\frac{x^{3}-3}{x^{11}}}dx\)


Use the sub \(\displaystyle \displaystyle x=u^{\frac{-1}{3}}, \;\ dx=\frac{-1}{3}u^{-\frac{4}{3}}du\)

This will whittle it down to something more workable. Be careful with the algebra. Specifically, the exponent laws.
 
Well I am lost still

u= 1-3/x^(-3) du=9x^(-4)dx then dx= (1/9)x^4


So this gets you nowhere??

{integral) ((1/9)x^(-4)u)^(1/2) dx


Now what? You can't go anywhere else?
 
Ok wait, I just saw your post Galactus. Now I will look at what you said...
 
Well, I am still lost on this one, I skipped ahead and did other problems hoping that I would come across some concept I am missing...

So Galactus,

this is the u, right?

1-3/x^(3) If it is, I don't even know how I would have come across that as the u unless I looked in the back of the book...

and this is embarrassing, I don't even know how you came up with the substitution x=u^(-1/3)

I am assuming u= 1-3/x^(3)

so I can only get as far as

x^(-3) = (1/3) - (U/3)

I am embarrassed that I don't know the algebra of how to get the x by itself from here. Is that what you were showing me, Galactus? Or am I going in a different direction?
 
50_93157d9b8020d9036663a8e140c765a8.png


I would use the substitution

\(\displaystyle x^3 \ = \ 3 * sec^2(\theta)\)
 
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