how do i solve this? thanks!!

- anonymous

how do i solve this? thanks!!

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- anonymous

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

- anonymous

what do you know so far?

- anonymous

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

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## More answers

- Astrophysics

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

- anonymous

you seem to know enough to tackle the problem

- Astrophysics

Try it out and see what you get!

- Astrophysics

You can also try \[u=\sqrt[3]{x}\] but then you probably will end up doing two substitutions, mess around with it and see what you get

- anonymous

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

- anonymous

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

- anonymous

oops *dx=3sqrtx ? :/

- Jhannybean

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

- Jhannybean

I left out the numbers

- anonymous

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

- ganeshie8

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

- anonymous

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

- ganeshie8

what good does that do

- anonymous

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

- ganeshie8

unless u hate the letter `x` ..

- zepdrix

\(\large\rm u=1+\sqrt[3]{x}\) or \(\large\rm u=\sqrt[3]{x}\)
I think both of these subs will work out just fine :)
Err no, go with the first one. I dunno bout that other guy, he gave me a funny look.

- zepdrix

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

- anonymous

lol im confused:/
so u = x ?

- Jhannybean

Oh what about changing this to arctan function..

- zepdrix

\[\large\rm u=1+\sqrt[3]{x},\qquad\qquad du=\frac{1}{3x^{2/3}}dx\]Do you understand how to differentiate that u? :o

- anonymous

i think so :) what is next?

- zepdrix

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

- anonymous

how do i do that?? :O

- anonymous

just differentiate what you have and try to finish it

- zepdrix

Isolate the dx, that's one of the things we're substituting stuff in for after all.\[\large\rm 3x^{2/3}du=dx\]I'm gonna use rules of exponents to write it like this:\[\large\rm 3(x^{1/3})^2du=dx\]

- anonymous

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

- zepdrix

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

- zepdrix

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

- zepdrix

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

- zepdrix

Was that super confusing? :o

- Jhannybean

Oh why did I miss this lol D'oh!

- anonymous

As I have said earlier, you have sufficient knowledge to tackle this problem. Just see to it that you finish what you started so you can learn different techniques along the way.

- anonymous

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

- zepdrix

Substitute in your pieces,\[\large\rm \color{orangered}{1+\sqrt[3]{x}=u},\qquad\qquad \color{royalblue}{dx=3(u-1)^2du}\]\[\large\rm \int\limits \frac{1}{\color{orangered}{1+\sqrt[3]{x}}}\color{royalblue}{dx}=?\]

- zepdrix

And from there, it shouldn't be too bad :)
You have to expand out the square in the numerator,
and you can divide each term by u, and integrate term by term.

- anonymous

best coaching ever

- anonymous

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

- anonymous

@zepdrix ?

- zepdrix

sec, doing some calculations :)

- anonymous

okie :)

- zepdrix

What did you get after integrating in u?
I feel like you're missing a 2 in the middle maybe.\[\large\rm =3\left(\frac{1}{2}u^2-2u+\ln u\right)\]Something similar or no? :o
We can substitute back in a sec, I'm just curious if yours looks the same up to this point.

- anonymous

yes, i have that :

- anonymous

:)

- zepdrix

When you get to this point, you normally have two options:
~find new boundary values for your integral, in terms of u, instead of x.
~or undo your substitution and use the original values for integrating.
I have a feeling... that with this monster it's going to be easier to get new boundaries for u as opposed to subbing back in those weird x's.\[\large\rm x=0 \qquad\to\qquad u=?\]\[\large\rm x=1\qquad\to\qquad u=?\]

- anonymous

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

- zepdrix

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

- anonymous

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

- anonymous

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

- zepdrix

Your ln(2) disappeared, uh oh

- anonymous

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

- zepdrix

You could go a tad further with it if you wanted, ln(1)=0.
\[\large\rm =ln(2^3)-\frac{3}{2}\]
but whatever,
yay good job \c:/

- anonymous

ooh okay, so would that be my solution?

- zepdrix

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

- anonymous

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

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