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

1. anonymous

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

2. anonymous

what do you know so far?

3. anonymous

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

4. Astrophysics

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

5. anonymous

you seem to know enough to tackle the problem

6. Astrophysics

Try it out and see what you get!

7. 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

8. anonymous

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

9. anonymous

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

10. anonymous

oops *dx=3sqrtx ? :/

11. Jhannybean

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

12. Jhannybean

I left out the numbers

13. anonymous

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

14. ganeshie8

you really want to sub $$u=x$$ ?

15. anonymous

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

16. ganeshie8

what good does that do

17. anonymous

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

18. ganeshie8

unless u hate the letter x ..

19. 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.

20. zepdrix

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

21. anonymous

lol im confused:/ so u = x ?

22. Jhannybean

Oh what about changing this to arctan function..

23. 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

24. anonymous

i think so :) what is next?

25. zepdrix

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

26. anonymous

how do i do that?? :O

27. anonymous

just differentiate what you have and try to finish it

28. 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:$\large\rm 3(x^{1/3})^2du=dx$

29. anonymous

that means $$\large x^\frac{1}{3}$$

30. zepdrix

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

31. zepdrix

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

32. zepdrix

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

33. zepdrix

Was that super confusing? :o

34. Jhannybean

Oh why did I miss this lol D'oh!

35. 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.

36. anonymous

okay, haha yes a bit confusing but i think i follow.. sort of…. :P what do i do next?

37. 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}=?$

38. 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.

39. anonymous

best coaching ever

40. anonymous

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

41. anonymous

@zepdrix ?

42. zepdrix

sec, doing some calculations :)

43. anonymous

okie :)

44. 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.

45. anonymous

yes, i have that :

46. anonymous

:)

47. 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=?$

48. anonymous

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

49. zepdrix

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

50. anonymous

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

51. anonymous

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

52. zepdrix

53. anonymous

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

54. 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:/

55. anonymous

ooh okay, so would that be my solution?

56. zepdrix

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

57. anonymous

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