anonymous
  • anonymous
how do i do this integral?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\int\limits_{} \frac{ 1 }{ \sqrt{x} - \sqrt[3]{x} }dx\]
anonymous
  • anonymous
and it says hint: substitute….\[u = \sqrt[6]{x}\]
anonymous
  • anonymous
not sure what i would be doing next? :/

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inkyvoyd
  • inkyvoyd
did you try it? what does du equal?
anonymous
  • anonymous
du = 1/6x^(5/6) ?
ganeshie8
  • ganeshie8
maybe this is an easier form : \(u = \sqrt[6]{x} \implies u^6 = x\) now try expressing \(dx\) in terms of \(du\)
anonymous
  • anonymous
okay, so dx = 6u^5 ?
ganeshie8
  • ganeshie8
Yes, keep going..
anonymous
  • anonymous
oh yes, oops :P what am i looking for next? :/
ganeshie8
  • ganeshie8
pretty sure you meant dx = 6u^5 `du`
anonymous
  • anonymous
do i need to find any other ones? :/
ganeshie8
  • ganeshie8
you want to evaluate the given antiderivative
anonymous
  • anonymous
so we get 30u^5 ?
anonymous
  • anonymous
not sure if I'm doing the right thing? :/
anonymous
  • anonymous
oops ^4
anonymous
  • anonymous
not sure, wait evaluate the antiderivative? so am i plugging in? :/
Astrophysics
  • Astrophysics
\[\frac{ dx }{ du }\] is what you want
anonymous
  • anonymous
okay, so i get 6u^5 / sqrt x - x^1/3 ?
Astrophysics
  • Astrophysics
\[x=u^6\]
anonymous
  • anonymous
so dx = 6u^5 du ? but how do i find du?
Jhannybean
  • Jhannybean
dx = 6u^5 `du` So if you treated du as a variable... how would you isolate it to one side?
anonymous
  • anonymous
ohh du = -6u^5 dx ?
Jhannybean
  • Jhannybean
Where did the -ve come from....
anonymous
  • anonymous
-ve?
anonymous
  • anonymous
hahaha ooh not sure so it is just positive?
anonymous
  • anonymous
so dx is just 1/6u^5 ?
anonymous
  • anonymous
ohhhh okay i see now :) oopsies sorry!! what happens next?
Astrophysics
  • Astrophysics
Ok I feel this is all over the place, so I'm just going to restart
Astrophysics
  • Astrophysics
\[u^6 = x \implies 6u^5du = dx\] so far so good?
anonymous
  • anonymous
yes:)
Astrophysics
  • Astrophysics
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?
anonymous
  • anonymous
okay!! we get -sqrt u^6/4 -3u^3/4 ? :/
Astrophysics
  • Astrophysics
I don't know what you did, but just simplify the integrand so we get \[6 \int\limits \frac{ u^5 }{ u^3-u^2 }du\]
Astrophysics
  • Astrophysics
Can you finish it off from there?
Astrophysics
  • Astrophysics
\[\int\limits \frac{ 6u^5 }{ u^{6/2}-u^{6/2} }du \implies 6 \int\limits\limits \frac{ u^5 }{ u^3-u^2 }du\] so it's more clear
anonymous
  • anonymous
hopefully :P would i get this? \[\frac{ u^3 }{ 3 } + \frac{ u^2 }{ 2 } + u + \log(u-1) +C \]
anonymous
  • anonymous
would this be my solution?
Astrophysics
  • Astrophysics
Yeah, but you have the 6 out there to, so you have to multiply through by 6 and subtitute your original \[u = \sqrt[6]{x}\]
anonymous
  • anonymous
ohhh okay, so i would get this? 2u^3 + 3u^2 + 6u + 6log(u-1) + C ?
anonymous
  • anonymous
and then wherever there is a u, i just plug that in and whatever i get there will be my final solution?
Astrophysics
  • Astrophysics
yes
anonymous
  • anonymous
yay!! thank you!!: )
Astrophysics
  • Astrophysics
Np
Astrophysics
  • Astrophysics
You should put absolute values ln|...|
anonymous
  • anonymous
okie!! :D
Jhannybean
  • Jhannybean
Can you show how you would simplify the integral, \[6\int \frac{u^5}{u^3-u^2}du\]
Jhannybean
  • Jhannybean
@iheartfood how did you go from the integral to \(\dfrac{ u^3 }{ 3 } + \dfrac{ u^2 }{ 2 } + u + \log(u-1) +C\) ?
anonymous
  • anonymous
i split them up into different parts!!
Jhannybean
  • Jhannybean
what do you mean?
Astrophysics
  • Astrophysics
Long division > +1-1 trick xd
Jhannybean
  • Jhannybean
\[\int \frac{u^5}{u^3-u^2} du =\int \frac{u^5}{u^2(u-1)}du =\int \frac{u^3}{u-1}du \]|dw:1442987647349:dw| \[\int \left(u^2+u+1+\frac{1}{u-1}\right)du\]\[=\frac{u^3}{3}+\frac{u^2}{2}+u+\log(u-1) +c\]
Astrophysics
  • Astrophysics
Yeah perfect! But the original question had a 6 outside, haha but great work!
Jhannybean
  • Jhannybean
\[=6\left[ \frac{u^3}{3}+\frac{u^2}{2}+u+\log(u-1)+c\right]\] \[=2u^3+3u^2+6u+\log(u-1)^6+6c\] this? lol
Astrophysics
  • Astrophysics
\[=2u^3+3u^2+6u+6\ln|u-1|+c\] looks good
Jhannybean
  • Jhannybean
Ahh.
Astrophysics
  • Astrophysics
Now we just plug our original substitution and we're done
Jhannybean
  • Jhannybean
Yeah.
Astrophysics
  • Astrophysics
The other method is you basically subtract and add 1, then separate the numerator so we have \[6 \int\limits \frac{ u^3-1+1 }{ u-1 }du \implies 6 \int\limits \left( \frac{ u^3-1 }{ u-1 }+\frac{ 1 }{ u-1 } \right)du\]
Astrophysics
  • Astrophysics
\[(a^3-b^3)=(a-b)(a^2-ab+2b^2)\]...etc
Astrophysics
  • Astrophysics
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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