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anonymous
 one year ago
how do i solve this? thanks!!
anonymous
 one year ago
how do i solve this? thanks!!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{1} \frac{ 1 }{ 1+\sqrt[3]{x} }dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what do you know so far?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would u = 1 + 3sqrtx and du = dx?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1This requires a u sub, what is the obvious u sub here?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you seem to know enough to tackle the problem

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Try it out and see what you get!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1You 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
 one year ago
Best ResponseYou've already chosen the best response.0ohh okay, err so would this be on the right track? u = 1 + 3sqrtx and du = 3udu du = 3sqrtx

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm still quite confused :( @Astrophysics :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops *dx=3sqrtx ? :/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int \frac{dx}{1+\sqrt[3]{x}}\]\[u=x~,~ du=dx\]\[\int\frac{du}{1+u^{1/3}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I left out the numbers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay!! and so the next step would be the solve this integral? :/

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1you really want to sub \(u=x\) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait, so we don't sub u=x? :/

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1what good does that do

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im not sure :/ what would i make the sub be then? :/

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1unless u hate the letter `x` ..

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.6\(\large\rm u=1+\sqrt[3]{x}\) 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
 one year ago
Best ResponseYou've already chosen the best response.6Where'd you leave off food man? :O Where you stuck at?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol im confused:/ so u = x ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh what about changing this to arctan function..

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.6\[\large\rm u=1+\sqrt[3]{x},\qquad\qquad du=\frac{1}{3x^{2/3}}dx\]Do you understand how to differentiate that u? :o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think so :) what is next?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.6Now you have to apply a bunch of sneaky little tricks!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how do i do that?? :O

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just differentiate what you have and try to finish it

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.6Isolate 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
 one year ago
Best ResponseYou've already chosen the best response.0that means \(\large x^\frac{1}{3} \)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.6We can sub something in for that x^(1/3) by using our equation involving u.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.6\(\large\rm u=1+x^{1/3}\quad\to\quad x^{1/3}=u1\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.6So there is our dx,\[\large\rm 3(u1)^2du=dx\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.6Was that super confusing? :o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh why did I miss this lol D'oh!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0As 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
 one year ago
Best ResponseYou've already chosen the best response.0okay, haha yes a bit confusing but i think i follow.. sort of…. :P what do i do next?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.6Substitute in your pieces,\[\large\rm \color{orangered}{1+\sqrt[3]{x}=u},\qquad\qquad \color{royalblue}{dx=3(u1)^2du}\]\[\large\rm \int\limits \frac{1}{\color{orangered}{1+\sqrt[3]{x}}}\color{royalblue}{dx}=?\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.6And 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
 one year ago
Best ResponseYou've already chosen the best response.0okay, er so do i get this? \[\frac{ 3x ^{2/3} }{ 2 }  3\sqrt[3]{x} + 3\log(\sqrt[3]{x}+1) + C \]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.6sec, doing some calculations :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.6What did you get after integrating in u? I feel like you're missing a 2 in the middle maybe.\[\large\rm =3\left(\frac{1}{2}u^22u+\ln u\right)\]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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.6When 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
 one year ago
Best ResponseYou've already chosen the best response.0ohh okay :) u=1 , u=2 ?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.6\[\large\rm =3\left(\frac{1}{2}u^22u+\ln u\right)_1^2\]Mmm sounds good!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so now i plug in? and i get 3((24+ln(2) (1/22+ln(1)) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 3(2+ln(2)  1/2 +2 ln(1) = 3(1/2 +ln(1) ?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.6Your ln(2) disappeared, uh oh

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh oops so it should be 3(1/2 + ln(2) + ln(1)) ?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.6You 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
 one year ago
Best ResponseYou've already chosen the best response.0ooh okay, so would that be my solution?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.6Yes, that or a decimal value, depending on which your teacher prefers.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooh yay!! thanks so much!! to all of you!!:)
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