anonymous
  • anonymous
how do i solve this? thanks!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\int\limits_{0}^{1} \frac{ 1 }{ 1+\sqrt[3]{x} }dx\]
anonymous
  • anonymous
what do you know so far?
anonymous
  • anonymous
would u = 1 + 3sqrtx and du = dx?

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Astrophysics
  • Astrophysics
This requires a u sub, what is the obvious u sub here?
anonymous
  • anonymous
you seem to know enough to tackle the problem
Astrophysics
  • Astrophysics
Try it out and see what you get!
Astrophysics
  • 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
  • anonymous
ohh okay, err so would this be on the right track? u = 1 + 3sqrtx and du = 3udu du = 3sqrtx
anonymous
  • anonymous
I'm still quite confused :( @Astrophysics :(
anonymous
  • anonymous
oops *dx=3sqrtx ? :/
Jhannybean
  • Jhannybean
\[\int \frac{dx}{1+\sqrt[3]{x}}\]\[u=x~,~ du=dx\]\[\int\frac{du}{1+u^{1/3}}\]
Jhannybean
  • Jhannybean
I left out the numbers
anonymous
  • anonymous
okay!! and so the next step would be the solve this integral? :/
ganeshie8
  • ganeshie8
you really want to sub \(u=x\) ?
anonymous
  • anonymous
wait, so we don't sub u=x? :/
ganeshie8
  • ganeshie8
what good does that do
anonymous
  • anonymous
im not sure :/ what would i make the sub be then? :/
ganeshie8
  • ganeshie8
unless u hate the letter `x` ..
zepdrix
  • 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
  • zepdrix
Where'd you leave off food man? :O Where you stuck at?
anonymous
  • anonymous
lol im confused:/ so u = x ?
Jhannybean
  • Jhannybean
Oh what about changing this to arctan function..
zepdrix
  • 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
  • anonymous
i think so :) what is next?
zepdrix
  • zepdrix
Now you have to apply a bunch of sneaky little tricks!
anonymous
  • anonymous
how do i do that?? :O
anonymous
  • anonymous
just differentiate what you have and try to finish it
zepdrix
  • 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
  • anonymous
that means \(\large x^\frac{1}{3} \)
zepdrix
  • zepdrix
We can sub something in for that x^(1/3) by using our equation involving u.
zepdrix
  • zepdrix
\(\large\rm u=1+x^{1/3}\quad\to\quad x^{1/3}=u-1\)
zepdrix
  • zepdrix
So there is our dx,\[\large\rm 3(u-1)^2du=dx\]
zepdrix
  • zepdrix
Was that super confusing? :o
Jhannybean
  • Jhannybean
Oh why did I miss this lol D'oh!
anonymous
  • 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
  • anonymous
okay, haha yes a bit confusing but i think i follow.. sort of…. :P what do i do next?
zepdrix
  • 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
  • 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
  • anonymous
best coaching ever
anonymous
  • 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
  • anonymous
@zepdrix ?
zepdrix
  • zepdrix
sec, doing some calculations :)
anonymous
  • anonymous
okie :)
zepdrix
  • 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
  • anonymous
yes, i have that :
anonymous
  • anonymous
:)
zepdrix
  • 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
  • anonymous
ohh okay :) u=1 , u=2 ?
zepdrix
  • zepdrix
\[\large\rm =3\left(\frac{1}{2}u^2-2u+\ln u\right)_1^2\]Mmm sounds good!
anonymous
  • anonymous
so now i plug in? and i get 3((2-4+ln(2) -(1/2-2+ln(1)) ?
anonymous
  • anonymous
so 3(-2+ln(2) - 1/2 +2 -ln(1) = 3(-1/2 +ln(1) ?
zepdrix
  • zepdrix
Your ln(2) disappeared, uh oh
anonymous
  • anonymous
oh oops so it should be 3(-1/2 + ln(2) + ln(1)) ?
zepdrix
  • 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
  • anonymous
ooh okay, so would that be my solution?
zepdrix
  • zepdrix
Yes, that or a decimal value, depending on which your teacher prefers.
anonymous
  • anonymous
ooh yay!! thanks so much!! to all of you!!:)

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