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\[\int\limits_{0}^{1} \frac{ 1 }{ 1+\sqrt[3]{x} }dx\]

what do you know so far?

would u = 1 + 3sqrtx and du = dx?

This requires a u sub, what is the obvious u sub here?

you seem to know enough to tackle the problem

Try it out and see what you get!

ohh okay, err so would this be on the right track?
u = 1 + 3sqrtx and du = 3udu du = 3sqrtx

I'm still quite confused :( @Astrophysics :(

oops *dx=3sqrtx ? :/

\[\int \frac{dx}{1+\sqrt[3]{x}}\]\[u=x~,~ du=dx\]\[\int\frac{du}{1+u^{1/3}}\]

I left out the numbers

okay!! and so the next step would be the solve this integral? :/

you really want to sub \(u=x\) ?

wait, so we don't sub u=x? :/

what good does that do

im not sure :/ what would i make the sub be then? :/

unless u hate the letter `x` ..

Where'd you leave off food man? :O
Where you stuck at?

lol im confused:/
so u = x ?

Oh what about changing this to arctan function..

i think so :) what is next?

Now you have to apply a bunch of sneaky little tricks!

how do i do that?? :O

just differentiate what you have and try to finish it

that means \(\large x^\frac{1}{3} \)

We can sub something in for that x^(1/3) by using our equation involving u.

\(\large\rm u=1+x^{1/3}\quad\to\quad x^{1/3}=u-1\)

So there is our dx,\[\large\rm 3(u-1)^2du=dx\]

Was that super confusing? :o

Oh why did I miss this lol D'oh!

okay, haha yes a bit confusing but i think i follow.. sort ofâ€¦. :P
what do i do next?

best coaching ever

okay, er so do i get this? \[\frac{ 3x ^{2/3} }{ 2 } - 3\sqrt[3]{x} + 3\log(\sqrt[3]{x}+1) + C \]

sec, doing some calculations :)

okie :)

yes, i have that :

:)

ohh okay :) u=1 , u=2 ?

\[\large\rm =3\left(\frac{1}{2}u^2-2u+\ln u\right)_1^2\]Mmm sounds good!

so now i plug in? and i get
3((2-4+ln(2) -(1/2-2+ln(1)) ?

so 3(-2+ln(2) - 1/2 +2 -ln(1) = 3(-1/2 +ln(1) ?

Your ln(2) disappeared, uh oh

oh oops so it should be 3(-1/2 + ln(2) + ln(1)) ?

ooh okay, so would that be my solution?

Yes, that or a decimal value, depending on which your teacher prefers.

ooh yay!! thanks so much!! to all of you!!:)