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anonymous

  • 4 years ago

more calculus help

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  1. anonymous
    • 4 years ago
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    \[\int\limits_{?}^{?}(1)/(xln(x ^{4}))\]

  2. bahrom7893
    • 4 years ago
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    that 1/x is buggin me as a u sub.. but it's prolly integration by parts

  3. anonymous
    • 4 years ago
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    start with \[\frac{1}{x\ln(x^4)}=\frac{1}{4x\ln(x)}\] then it should be easy

  4. bahrom7893
    • 4 years ago
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    Let u = ln(x^4)

  5. bahrom7893
    • 4 years ago
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    Ha I think i got this one!

  6. bahrom7893
    • 4 years ago
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    u = ln(x^4) du = 4/x simple u sub!

  7. anonymous
    • 4 years ago
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    good. because i am clueless???

  8. anonymous
    • 4 years ago
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    pull the 1/4 outside of the integral get \[\frac{1}{4}\int \frac{dx}{x\ln(x)}\] then make \[u=\ln(x),du=\frac{1}{x}dx\] and you are home free

  9. bahrom7893
    • 4 years ago
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    well yea pretty much the same thing as satellite did

  10. anonymous
    • 4 years ago
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    don't forget the properties of the log!

  11. bahrom7893
    • 4 years ago
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    lol my method works too though :)

  12. anonymous
    • 4 years ago
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    oookkk i got it

  13. anonymous
    • 4 years ago
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    yes it will work and you will see that if \[u=\ln(x^4)\] then \[du=\frac{4}{x}dx\] but that is telling you that \[\ln(x^4)=4\ln(x)\]

  14. anonymous
    • 4 years ago
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    would the fianl answer be 1/4ln (ln(x))+c?

  15. anonymous
    • 4 years ago
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    Final*

  16. anonymous
    • 4 years ago
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    it would be, yes

  17. anonymous
    • 4 years ago
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    sweet thanks again.

  18. anonymous
    • 4 years ago
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    btw if you notice you will get a different answer from wolfram and if you like i can explain why

  19. anonymous
    • 4 years ago
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    please do

  20. anonymous
    • 4 years ago
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    here is what wolfram writes if you just type it in. you get \[\frac{1}{4}\ln(\ln(x^4))+c\]

  21. anonymous
    • 4 years ago
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    but \[\ln(\ln(x^4))=\ln(4\ln(x))=\ln(4)+\ln(\ln(x))\] and \[\ln(4) \] is a constant. so answers are the same, since the constant is just a constant, like the +C out at the end

  22. anonymous
    • 4 years ago
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    oo ok that is simple enough thanks again

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