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koreanub

  • 3 years ago

Integration

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  1. koreanub
    • 3 years ago
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    \[\int\limits_{?}^{?} \frac{ 1 }{ \sqrt{x}(1+x) }\]

  2. koreanub
    • 3 years ago
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    The question marks are not suppose to be there, so it's the problem without the question marks.

  3. experimentX
    • 3 years ago
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    \int just use like this for integration ... seems like trig subs seems to work also changing 1 + x = (1 + i sqrt(x))(1 - i sqrt(x)) and then taking partial fraction might work.

  4. mukushla
    • 3 years ago
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    i believe u-sub will work

  5. mukushla
    • 3 years ago
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    \[u=\sqrt{x}\] ha?

  6. experimentX
    • 3 years ago
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    woops!! that also works!!

  7. koreanub
    • 3 years ago
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    For this exercise, I'm technically not suposed to use partial fractions

  8. koreanub
    • 3 years ago
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    I may be missing something, but I can't seem to find the answer with u = \[\sqrt{x} \]

  9. experimentX
    • 3 years ago
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    change all x's into u's probably you would end up with 1/1+u^2 or something like that.

  10. zepdrix
    • 3 years ago
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    You have to mess with the U a little bit kore :) but it will work.

  11. hartnn
    • 3 years ago
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    x= tan^2 theta might work

  12. experimentX
    • 3 years ago
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    this is same as using trig subs x = tan^2 theta in original.

  13. experimentX
    • 3 years ago
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    but lol ... we know int 1/1+x^2 = arctan (x) which is one of the standard integral.

  14. koreanub
    • 3 years ago
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    I got it! Yes!!! I was substituting the first \[\sqrt{x} \] in the integrand. Then, I realized that it gets canceled out. Thank you very much everyone! ... If it wasn't for you, I probably would have spent another 10 minutes on this. -.-

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