Question

how do i do this integral?

Answer 1

\[\int\limits_{} \frac{ 1 }{ \sqrt{x} - \sqrt[3]{x} }dx\]

Answer 2

and it says hint: substitute….\[u = \sqrt[6]{x}\]

Answer 3

not sure what i would be doing next? :/

Answer 4

did you try it? what does du equal?

Answer 5

du = 1/6x^(5/6) ?

Answer 6

maybe this is an easier form :
\(u = \sqrt[6]{x} \implies u^6 = x\)

now try expressing \(dx\) in terms of \(du\)

Answer 7

okay, so dx = 6u^5 ?

Answer 8

Yes, keep going..

Answer 9

oh yes, oops :P what am i looking for next? :/

Answer 10

pretty sure you meant
dx = 6u^5 `du`

Answer 11

do i need to find any other ones? :/

Answer 12

you want to evaluate the given antiderivative

Answer 13

so we get 30u^5 ?

Answer 14

not sure if I'm doing the right thing? :/

Answer 15

oops ^4

Answer 16

not sure, wait evaluate the antiderivative? so am i plugging in? :/

Answer 17

\[\frac{ dx }{ du }\] is what you want

Answer 18

okay, so i get 6u^5 / sqrt x - x^1/3 ?

Answer 19

\[x=u^6\]

Answer 20

so dx = 6u^5 du ?

but how do i find du?

Answer 21

dx = 6u^5 `du`

So if you treated du as a variable... how would you isolate it to one side?

Answer 22

ohh du = -6u^5 dx ?

Answer 23

Where did the -ve come from....

Answer 24

-ve?

Answer 25

hahaha ooh not sure so it is just positive?

Answer 26

so dx is just 1/6u^5 ?

Answer 27

ohhhh okay i see now :) oopsies sorry!! what happens next?

Answer 28

Ok I feel this is all over the place, so I'm just going to restart

Answer 29

\[u^6 = x \implies 6u^5du = dx\] so far so good?

Answer 30

yes:)

Answer 31

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?

Answer 32

okay!! we get -sqrt u^6/4 -3u^3/4 ? :/

Answer 33

I don't know what you did, but just simplify the integrand so we get \[6 \int\limits \frac{ u^5 }{ u^3-u^2 }du\]

Answer 34

Can you finish it off from there?

Answer 35

\[\int\limits \frac{ 6u^5 }{ u^{6/2}-u^{6/2} }du \implies 6 \int\limits\limits \frac{ u^5 }{ u^3-u^2 }du\] so it's more clear

Answer 36

hopefully :P

would i get this? \[\frac{ u^3 }{ 3 } + \frac{ u^2 }{ 2 } + u + \log(u-1) +C \]

Answer 37

would this be my solution?

Answer 38

Yeah, but you have the 6 out there to, so you have to multiply through by 6 and subtitute your original \[u = \sqrt[6]{x}\]

Answer 39

ohhh okay, so i would get this?

2u^3 + 3u^2 + 6u + 6log(u-1) + C ?

Answer 40

and then wherever there is a u, i just plug that in and whatever i get there will be my final solution?

Answer 41

yes

Answer 42

yay!! thank you!!: )

Answer 43

Np

Answer 44

You should put absolute values ln|...|

Answer 45

okie!! :D

Answer 46

Can you show how you would simplify the integral, \[6\int \frac{u^5}{u^3-u^2}du\]

Answer 47

@iheartfood how did you go from the integral to \(\dfrac{ u^3 }{ 3 } + \dfrac{ u^2 }{ 2 } + u + \log(u-1) +C\) ?

Answer 48

i split them up into different parts!!

Answer 49

what do you mean?

Answer 50

Long division > +1-1 trick xd

Answer 51

\[\int \frac{u^5}{u^3-u^2} du =\int \frac{u^5}{u^2(u-1)}du =\int \frac{u^3}{u-1}du \] \[\int \left(u^2+u+1+\frac{1}{u-1}\right)du\]\[=\frac{u^3}{3}+\frac{u^2}{2}+u+\log(u-1) +c\]