how do i do this integral?

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how do i do this integral?

Mathematics
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\[\int\limits_{} \frac{ 1 }{ \sqrt{x} - \sqrt[3]{x} }dx\]
and it says hint: substitute….\[u = \sqrt[6]{x}\]
not sure what i would be doing next? :/

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Other answers:

did you try it? what does du equal?
du = 1/6x^(5/6) ?
maybe this is an easier form : \(u = \sqrt[6]{x} \implies u^6 = x\) now try expressing \(dx\) in terms of \(du\)
okay, so dx = 6u^5 ?
Yes, keep going..
oh yes, oops :P what am i looking for next? :/
pretty sure you meant dx = 6u^5 `du`
do i need to find any other ones? :/
you want to evaluate the given antiderivative
so we get 30u^5 ?
not sure if I'm doing the right thing? :/
oops ^4
not sure, wait evaluate the antiderivative? so am i plugging in? :/
\[\frac{ dx }{ du }\] is what you want
okay, so i get 6u^5 / sqrt x - x^1/3 ?
\[x=u^6\]
so dx = 6u^5 du ? but how do i find du?
dx = 6u^5 `du` So if you treated du as a variable... how would you isolate it to one side?
ohh du = -6u^5 dx ?
Where did the -ve come from....
-ve?
hahaha ooh not sure so it is just positive?
so dx is just 1/6u^5 ?
ohhhh okay i see now :) oopsies sorry!! what happens next?
Ok I feel this is all over the place, so I'm just going to restart
\[u^6 = x \implies 6u^5du = dx\] so far so good?
yes:)
Now don't start moving things to dx, because that doesn't make much sense, so lets just sub this into our integral now...\[\int\limits \frac{ 6u^5du }{ \sqrt{u^6}-\sqrt[3]{u^6} }\] now simplify this, what do you get?
okay!! we get -sqrt u^6/4 -3u^3/4 ? :/
I don't know what you did, but just simplify the integrand so we get \[6 \int\limits \frac{ u^5 }{ u^3-u^2 }du\]
Can you finish it off from there?
\[\int\limits \frac{ 6u^5 }{ u^{6/2}-u^{6/2} }du \implies 6 \int\limits\limits \frac{ u^5 }{ u^3-u^2 }du\] so it's more clear
hopefully :P would i get this? \[\frac{ u^3 }{ 3 } + \frac{ u^2 }{ 2 } + u + \log(u-1) +C \]
would this be my solution?
Yeah, but you have the 6 out there to, so you have to multiply through by 6 and subtitute your original \[u = \sqrt[6]{x}\]
ohhh okay, so i would get this? 2u^3 + 3u^2 + 6u + 6log(u-1) + C ?
and then wherever there is a u, i just plug that in and whatever i get there will be my final solution?
yes
yay!! thank you!!: )
Np
You should put absolute values ln|...|
okie!! :D
Can you show how you would simplify the integral, \[6\int \frac{u^5}{u^3-u^2}du\]
@iheartfood how did you go from the integral to \(\dfrac{ u^3 }{ 3 } + \dfrac{ u^2 }{ 2 } + u + \log(u-1) +C\) ?
i split them up into different parts!!
what do you mean?
Long division > +1-1 trick xd
\[\int \frac{u^5}{u^3-u^2} du =\int \frac{u^5}{u^2(u-1)}du =\int \frac{u^3}{u-1}du \]|dw:1442987647349:dw| \[\int \left(u^2+u+1+\frac{1}{u-1}\right)du\]\[=\frac{u^3}{3}+\frac{u^2}{2}+u+\log(u-1) +c\]
Yeah perfect! But the original question had a 6 outside, haha but great work!
\[=6\left[ \frac{u^3}{3}+\frac{u^2}{2}+u+\log(u-1)+c\right]\] \[=2u^3+3u^2+6u+\log(u-1)^6+6c\] this? lol
\[=2u^3+3u^2+6u+6\ln|u-1|+c\] looks good
Ahh.
Now we just plug our original substitution and we're done
Yeah.
The other method is you basically subtract and add 1, then separate the numerator so we have \[6 \int\limits \frac{ u^3-1+1 }{ u-1 }du \implies 6 \int\limits \left( \frac{ u^3-1 }{ u-1 }+\frac{ 1 }{ u-1 } \right)du\]
\[(a^3-b^3)=(a-b)(a^2-ab+2b^2)\]...etc
Guess I'll show it xD \[6 \int\limits \left( \frac{ (u-1)(u^2+u+1) }{ (u-1) }+\frac{ 1 }{ u-1 } \right)du \implies 6 \int\limits (u^2+u+1+\frac{ 1 }{ u-1 }) du\]

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