how do i solve this? thanks!!

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how do i solve this? thanks!!

Mathematics
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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?

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Other answers:

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!
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
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` ..
\(\large\rm u=1+\sqrt[3]{x}\) 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.
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..
\[\large\rm u=1+\sqrt[3]{x},\qquad\qquad du=\frac{1}{3x^{2/3}}dx\]Do you understand how to differentiate that u? :o
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
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:\[\large\rm 3(x^{1/3})^2du=dx\]
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!
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.
okay, haha yes a bit confusing but i think i follow.. sort of…. :P what do i do next?
Substitute in your pieces,\[\large\rm \color{orangered}{1+\sqrt[3]{x}=u},\qquad\qquad \color{royalblue}{dx=3(u-1)^2du}\]\[\large\rm \int\limits \frac{1}{\color{orangered}{1+\sqrt[3]{x}}}\color{royalblue}{dx}=?\]
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.
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 :)
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.
yes, i have that :
:)
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.\[\large\rm x=0 \qquad\to\qquad u=?\]\[\large\rm x=1\qquad\to\qquad u=?\]
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)) ?
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:/
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!!:)

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