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integral of 1/(u^3) ?

Mathematics
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@zzr0ck3r u there?
\[\large \int\limits \frac{ 1 }{ u^3 }du= \int\limits u^{-3}du\] you know how to integrate that right.
to tell you the truth I already knew how to do it, I just think this exersise is wrong so I need to verify, look at the bottom http://screencast.com/t/XEJ9Dmj1S

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Other answers:

looks good
my result: (1/2)*((-1/18) + (1/72))
what is the problem?
http://screencast.com/t/XEJ9Dmj1S this solution displays a different result
Just making sure I didnt miss anything lol
where did the \(\frac{1}{18}\) come from ?
the solution looks fine
3^2 = 9*2 = 18
show us what you did, did u change your bounds when you substituted?
3^2 = 3*3 = 9
three squared is not equal to nine squared
9*2(denominator) = 18
18 = 2*3^2
Right
I m confused:)
;p
\[F(u)=-\frac{1}{4u^2}\] \[F(3)-F(6)=-\frac{1}{4}\left(\frac{1}{9}-\frac{1}{36}\right)\]
where did the 1/4 come from
there might be a probability that my solution and the other one are equal idk
:D
the u - sub gave a factor of \(\frac{1}{2}\) and the anti derivative gave another \(-\frac{1}{2}\)
aah!
hold on
That was the trick, thank you
yw

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