## anonymous one year ago how do i do this integral?

1. anonymous

$\int\limits_{} \frac{ 1 }{ \sqrt{x} - \sqrt[3]{x} }dx$

2. anonymous

and it says hint: substitute….$u = \sqrt[6]{x}$

3. anonymous

not sure what i would be doing next? :/

4. inkyvoyd

did you try it? what does du equal?

5. anonymous

du = 1/6x^(5/6) ?

6. ganeshie8

maybe this is an easier form : $$u = \sqrt[6]{x} \implies u^6 = x$$ now try expressing $$dx$$ in terms of $$du$$

7. anonymous

okay, so dx = 6u^5 ?

8. ganeshie8

Yes, keep going..

9. anonymous

oh yes, oops :P what am i looking for next? :/

10. ganeshie8

pretty sure you meant dx = 6u^5 du

11. anonymous

do i need to find any other ones? :/

12. ganeshie8

you want to evaluate the given antiderivative

13. anonymous

so we get 30u^5 ?

14. anonymous

not sure if I'm doing the right thing? :/

15. anonymous

oops ^4

16. anonymous

not sure, wait evaluate the antiderivative? so am i plugging in? :/

17. Astrophysics

$\frac{ dx }{ du }$ is what you want

18. anonymous

okay, so i get 6u^5 / sqrt x - x^1/3 ?

19. Astrophysics

$x=u^6$

20. anonymous

so dx = 6u^5 du ? but how do i find du?

21. anonymous

dx = 6u^5 du So if you treated du as a variable... how would you isolate it to one side?

22. anonymous

ohh du = -6u^5 dx ?

23. anonymous

Where did the -ve come from....

24. anonymous

-ve?

25. anonymous

hahaha ooh not sure so it is just positive?

26. anonymous

so dx is just 1/6u^5 ?

27. anonymous

ohhhh okay i see now :) oopsies sorry!! what happens next?

28. Astrophysics

Ok I feel this is all over the place, so I'm just going to restart

29. Astrophysics

$u^6 = x \implies 6u^5du = dx$ so far so good?

30. anonymous

yes:)

31. Astrophysics

Now don't start moving things to dx, because that doesn't make much sense, so lets just sub this into our integral now...$\int\limits \frac{ 6u^5du }{ \sqrt{u^6}-\sqrt[3]{u^6} }$ now simplify this, what do you get?

32. anonymous

okay!! we get -sqrt u^6/4 -3u^3/4 ? :/

33. Astrophysics

I don't know what you did, but just simplify the integrand so we get $6 \int\limits \frac{ u^5 }{ u^3-u^2 }du$

34. Astrophysics

Can you finish it off from there?

35. Astrophysics

$\int\limits \frac{ 6u^5 }{ u^{6/2}-u^{6/2} }du \implies 6 \int\limits\limits \frac{ u^5 }{ u^3-u^2 }du$ so it's more clear

36. anonymous

hopefully :P would i get this? $\frac{ u^3 }{ 3 } + \frac{ u^2 }{ 2 } + u + \log(u-1) +C$

37. anonymous

would this be my solution?

38. Astrophysics

Yeah, but you have the 6 out there to, so you have to multiply through by 6 and subtitute your original $u = \sqrt[6]{x}$

39. anonymous

ohhh okay, so i would get this? 2u^3 + 3u^2 + 6u + 6log(u-1) + C ?

40. anonymous

and then wherever there is a u, i just plug that in and whatever i get there will be my final solution?

41. Astrophysics

yes

42. anonymous

yay!! thank you!!: )

43. Astrophysics

Np

44. Astrophysics

You should put absolute values ln|...|

45. anonymous

okie!! :D

46. anonymous

Can you show how you would simplify the integral, $6\int \frac{u^5}{u^3-u^2}du$

47. anonymous

@iheartfood how did you go from the integral to $$\dfrac{ u^3 }{ 3 } + \dfrac{ u^2 }{ 2 } + u + \log(u-1) +C$$ ?

48. anonymous

i split them up into different parts!!

49. anonymous

what do you mean?

50. Astrophysics

Long division > +1-1 trick xd

51. anonymous

$\int \frac{u^5}{u^3-u^2} du =\int \frac{u^5}{u^2(u-1)}du =\int \frac{u^3}{u-1}du$|dw:1442987647349:dw| $\int \left(u^2+u+1+\frac{1}{u-1}\right)du$$=\frac{u^3}{3}+\frac{u^2}{2}+u+\log(u-1) +c$

52. Astrophysics

Yeah perfect! But the original question had a 6 outside, haha but great work!

53. anonymous

$=6\left[ \frac{u^3}{3}+\frac{u^2}{2}+u+\log(u-1)+c\right]$ $=2u^3+3u^2+6u+\log(u-1)^6+6c$ this? lol

54. Astrophysics

$=2u^3+3u^2+6u+6\ln|u-1|+c$ looks good

55. anonymous

Ahh.

56. Astrophysics

Now we just plug our original substitution and we're done

57. anonymous

Yeah.

58. Astrophysics

The other method is you basically subtract and add 1, then separate the numerator so we have $6 \int\limits \frac{ u^3-1+1 }{ u-1 }du \implies 6 \int\limits \left( \frac{ u^3-1 }{ u-1 }+\frac{ 1 }{ u-1 } \right)du$

59. Astrophysics

$(a^3-b^3)=(a-b)(a^2-ab+2b^2)$...etc

60. Astrophysics

Guess I'll show it xD $6 \int\limits \left( \frac{ (u-1)(u^2+u+1) }{ (u-1) }+\frac{ 1 }{ u-1 } \right)du \implies 6 \int\limits (u^2+u+1+\frac{ 1 }{ u-1 }) du$