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jennychan12

  • 3 years ago

antiderivative of (x^2+4x)^1/3 ?

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  1. jennychan12
    • 3 years ago
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    \[\sqrt[3]{x^2+4x}\]

  2. jennychan12
    • 3 years ago
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    i know that u = x^2+4x so du = (2x+4)dx where do i go from there?

  3. jennychan12
    • 3 years ago
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    @agent0smith @Dido525

  4. AtiFS
    • 3 years ago
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    well u multipl that by 1/3 too

  5. AtiFS
    • 3 years ago
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    1/3 dx (x^2 + 4x) = 1/3* (2x + 4)..I think so

  6. jennychan12
    • 3 years ago
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    how did you get that? can u explain?

  7. jennychan12
    • 3 years ago
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    cuz du = (2x+4)dx = (2(x+2))dx so 1/2du = (x+2)dx then...?

  8. AtiFS
    • 3 years ago
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    actually, the value of power multplies..hmm du = 1/3*(2x + 4)dx => 3*du = (2x + 4)dx

  9. zepdrix
    • 3 years ago
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    I'm pretty sure you need to either do a Trig-Sub, or Integration-By-Parts might also be possible. No U-Sub available though :( Have you learned either of those methods yet?

  10. jennychan12
    • 3 years ago
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    yeah the substitution method.

  11. zepdrix
    • 3 years ago
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    Trig subs? :D

  12. jennychan12
    • 3 years ago
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    i think so? remind me what that is?

  13. Kainui
    • 3 years ago
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    I think you might want to add and subtract +4 and -4 under the cube root so that you can factor.

  14. AtiFS
    • 3 years ago
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    google the concept behind it :)

  15. beginnersmind
    • 3 years ago
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    idk how to do this, I'm just posting so that I get notified when someone answers it.

  16. Kainui
    • 3 years ago
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    I could reduce it down so that it becomes \[2^{3/5}\int\limits_{}^{}\tan^{3/5} \theta \sec \theta d \theta\] after making the substitution: \[(x+2)=2\sec \theta\]

  17. Kainui
    • 3 years ago
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    the fraction should be 5/3 not 3/5 on the 2 and tan(theta)

  18. AtiFS
    • 3 years ago
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    this link will help you :) http://www.the-mathroom.ca/freebs/cali3/cali3.htm

  19. Kainui
    • 3 years ago
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    I can't seem to crack it. You're in calculus 1 or 2?

  20. jennychan12
    • 3 years ago
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    1

  21. jennychan12
    • 3 years ago
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    where did the tangent, secant stiff come from?

  22. zepdrix
    • 3 years ago
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    Ah sorry I was busy :C Kanoo been helping you? c:

  23. Kainui
    • 3 years ago
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    I was doing trigonometric substitution, but you don't learn that until cal 2. If there's an easier way with just using u-sub then I would be surprised! Good luck zeppy! =P

  24. zepdrix
    • 3 years ago
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    Ya I dunno sorry Jenny :c This doesn't seem like a problem you should be doing yet, hmm.

  25. Kainui
    • 3 years ago
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    Check this out: http://www.wolframalpha.com/input/?i=(x%5E2%2B4x)%5E1%2F3dx&t=crmtb01 I've never even heard of that stuff. Hypergeometric function? lol

  26. zepdrix
    • 3 years ago
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    Woah that looks fun :O

  27. jennychan12
    • 3 years ago
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    oy.....! -_-

  28. cinar
    • 3 years ago
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    |dw:1355977140299:dw|

  29. cinar
    • 3 years ago
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    |dw:1355977229842:dw|

  30. cinar
    • 3 years ago
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    |dw:1355977363994:dw|

  31. cinar
    • 3 years ago
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    the last sub. should be u=2cosht

  32. cinar
    • 3 years ago
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    @satellite73 @Zarkon

  33. cinar
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=%28sinht%29^%285%2F3%29dt

  34. cinar
    • 3 years ago
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    I do not have enough information to solve this question..

  35. jennychan12
    • 3 years ago
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    g is the antiderivative function if g(5) = 7, then g(1) = ?

  36. cinar
    • 3 years ago
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    why are giving this info now, you should mention this at the beginning.. come on..

  37. cinar
    • 3 years ago
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    |dw:1355996811040:dw|

  38. cinar
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=integrate+%28x^2%2B4x%29^%281%2F3%29dx+from+x%3D1+to+5

  39. jennychan12
    • 3 years ago
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    thanks for trying to help. :)

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