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anonymous

  • 5 years ago

how do you integrate (lnx)^2?

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  1. anonymous
    • 5 years ago
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    \[\int\limits_{?}^{?} (lnx)^{2}dx\]

  2. anonymous
    • 5 years ago
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    The answer to the integral would be x((ln(x))^2-2ln(x)+2) It's not a very fun integral :P

  3. anonymous
    • 5 years ago
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    i got that from wolfram too but i need to know how

  4. anonymous
    • 5 years ago
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    I think you treat it like chain rule...? Soo this integral is x(ln(x)^2-2ln(x)+2) and if it were the integral of ln(x)^3 it would be x(ln(x)^3-3ln(x)^2+6ln(x)-6) and so on and so forth.

  5. anonymous
    • 5 years ago
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    chain rule is from differentiating..the techniques i know are u sub, by parts, trig id, partial fraction...

  6. anonymous
    • 5 years ago
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    i know integrating lnx gives xlnx-x+C

  7. anonymous
    • 5 years ago
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    I have no idea how to explain it then xD I just know it. This is one thing i'm just not sure on how to explain. it's just ln(x)-1, ln(x)^2-2ln(x)+2, etc. It's just how i've known it. i guess it's never really been explained to me, its just how it is. Sorry i couldn't be of help :(

  8. anonymous
    • 5 years ago
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    well x*all those, but you get my point ha

  9. anonymous
    • 5 years ago
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    ok, just like idk why lnx is xlnx-x+C...fair enough

  10. anonymous
    • 5 years ago
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    o wait, cuz of int by parts...

  11. anonymous
    • 5 years ago
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    ln(x) is x*(ln(x)-1) because it's the negative number of whatever the most recent ln is.

  12. anonymous
    • 5 years ago
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    o it was by parts the whole time...got it

  13. TuringTest
    • 5 years ago
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    yes, by parts u=(lnx)^2 dv=1

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