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anonymous

  • one year ago

how do i do this integral?

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  1. anonymous
    • one year ago
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    \[\int\limits_{} \frac{ 1 }{ \sqrt{x} - \sqrt[3]{x} }dx\]

  2. anonymous
    • one year ago
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    and it says hint: substitute….\[u = \sqrt[6]{x}\]

  3. anonymous
    • one year ago
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    not sure what i would be doing next? :/

  4. inkyvoyd
    • one year ago
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    did you try it? what does du equal?

  5. anonymous
    • one year ago
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    du = 1/6x^(5/6) ?

  6. ganeshie8
    • one year ago
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    maybe this is an easier form : \(u = \sqrt[6]{x} \implies u^6 = x\) now try expressing \(dx\) in terms of \(du\)

  7. anonymous
    • one year ago
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    okay, so dx = 6u^5 ?

  8. ganeshie8
    • one year ago
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    Yes, keep going..

  9. anonymous
    • one year ago
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    oh yes, oops :P what am i looking for next? :/

  10. ganeshie8
    • one year ago
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    pretty sure you meant dx = 6u^5 `du`

  11. anonymous
    • one year ago
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    do i need to find any other ones? :/

  12. ganeshie8
    • one year ago
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    you want to evaluate the given antiderivative

  13. anonymous
    • one year ago
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    so we get 30u^5 ?

  14. anonymous
    • one year ago
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    not sure if I'm doing the right thing? :/

  15. anonymous
    • one year ago
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    oops ^4

  16. anonymous
    • one year ago
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    not sure, wait evaluate the antiderivative? so am i plugging in? :/

  17. Astrophysics
    • one year ago
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    \[\frac{ dx }{ du }\] is what you want

  18. anonymous
    • one year ago
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    okay, so i get 6u^5 / sqrt x - x^1/3 ?

  19. Astrophysics
    • one year ago
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    \[x=u^6\]

  20. anonymous
    • one year ago
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    so dx = 6u^5 du ? but how do i find du?

  21. Jhannybean
    • one year ago
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    dx = 6u^5 `du` So if you treated du as a variable... how would you isolate it to one side?

  22. anonymous
    • one year ago
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    ohh du = -6u^5 dx ?

  23. Jhannybean
    • one year ago
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    Where did the -ve come from....

  24. anonymous
    • one year ago
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    -ve?

  25. anonymous
    • one year ago
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    hahaha ooh not sure so it is just positive?

  26. anonymous
    • one year ago
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    so dx is just 1/6u^5 ?

  27. anonymous
    • one year ago
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    ohhhh okay i see now :) oopsies sorry!! what happens next?

  28. Astrophysics
    • one year ago
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    Ok I feel this is all over the place, so I'm just going to restart

  29. Astrophysics
    • one year ago
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    \[u^6 = x \implies 6u^5du = dx\] so far so good?

  30. anonymous
    • one year ago
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    yes:)

  31. Astrophysics
    • one year ago
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    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
    • one year ago
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    okay!! we get -sqrt u^6/4 -3u^3/4 ? :/

  33. Astrophysics
    • one year ago
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    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
    • one year ago
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    Can you finish it off from there?

  35. Astrophysics
    • one year ago
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    \[\int\limits \frac{ 6u^5 }{ u^{6/2}-u^{6/2} }du \implies 6 \int\limits\limits \frac{ u^5 }{ u^3-u^2 }du\] so it's more clear

  36. anonymous
    • one year ago
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    hopefully :P would i get this? \[\frac{ u^3 }{ 3 } + \frac{ u^2 }{ 2 } + u + \log(u-1) +C \]

  37. anonymous
    • one year ago
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    would this be my solution?

  38. Astrophysics
    • one year ago
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    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
    • one year ago
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    ohhh okay, so i would get this? 2u^3 + 3u^2 + 6u + 6log(u-1) + C ?

  40. anonymous
    • one year ago
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    and then wherever there is a u, i just plug that in and whatever i get there will be my final solution?

  41. Astrophysics
    • one year ago
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    yes

  42. anonymous
    • one year ago
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    yay!! thank you!!: )

  43. Astrophysics
    • one year ago
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    Np

  44. Astrophysics
    • one year ago
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    You should put absolute values ln|...|

  45. anonymous
    • one year ago
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    okie!! :D

  46. Jhannybean
    • one year ago
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    Can you show how you would simplify the integral, \[6\int \frac{u^5}{u^3-u^2}du\]

  47. Jhannybean
    • one year ago
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    @iheartfood how did you go from the integral to \(\dfrac{ u^3 }{ 3 } + \dfrac{ u^2 }{ 2 } + u + \log(u-1) +C\) ?

  48. anonymous
    • one year ago
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    i split them up into different parts!!

  49. Jhannybean
    • one year ago
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    what do you mean?

  50. Astrophysics
    • one year ago
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    Long division > +1-1 trick xd

  51. Jhannybean
    • one year ago
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    \[\int \frac{u^5}{u^3-u^2} du =\int \frac{u^5}{u^2(u-1)}du =\int \frac{u^3}{u-1}du \]|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
    • one year ago
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    Yeah perfect! But the original question had a 6 outside, haha but great work!

  53. Jhannybean
    • one year ago
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    \[=6\left[ \frac{u^3}{3}+\frac{u^2}{2}+u+\log(u-1)+c\right]\] \[=2u^3+3u^2+6u+\log(u-1)^6+6c\] this? lol

  54. Astrophysics
    • one year ago
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    \[=2u^3+3u^2+6u+6\ln|u-1|+c\] looks good

  55. Jhannybean
    • one year ago
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    Ahh.

  56. Astrophysics
    • one year ago
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    Now we just plug our original substitution and we're done

  57. Jhannybean
    • one year ago
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    Yeah.

  58. Astrophysics
    • one year ago
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    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
    • one year ago
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    \[(a^3-b^3)=(a-b)(a^2-ab+2b^2)\]...etc

  60. Astrophysics
    • one year ago
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    Guess I'll show it xD \[6 \int\limits \left( \frac{ (u-1)(u^2+u+1) }{ (u-1) }+\frac{ 1 }{ u-1 } \right)du \implies 6 \int\limits (u^2+u+1+\frac{ 1 }{ u-1 }) du\]

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