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Math Help - integration by parts''.

  1. #1
    Member i_zz_y_ill's Avatar
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    integration by parts''.

    integrate by parts y^2(log(y)) between 1 and x thats int,y^2(multiplie by logy) between 1 and x in the form A(x)+{intB(y) between 1 and xwith respect to y},,,,,,,,,,,,the answer is ((x^3)/3)multiplied by(logx)-((y^2)/2),,,,,,,sorry i dont quite have latex prrammed in my brain at the moment i will in the next few weeks,,if somebody has the method for this it wuld be much appreciated within the next 10 hours if not after,,,thanks and sorry for the inconvenience of presentation.
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  2. #2
    Super Member
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    Use integration by parts with u = \log{y}~~dv = y^2~dy.
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  3. #3
    Member i_zz_y_ill's Avatar
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    hmm thanks but what does du = with respect to y then??,,,,i looked it up and on yahoo answersit gave natural logs whihc isnt the answer oddly!!! hard question i think!,,,:P?!?
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    I think you do really need to read my signature to improve your partial integration, hence, when gettin' this is your head, you'll easily see what is the best election.

    \int{y^{2}\ln y\,dy}=\frac{1}{3}\int{\left( y^{3} \right)'\ln y\,dy}=\frac{1}{3}\left( y^{3}\ln y-\int{y^{2}\,dy} \right).
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  5. #5
    Super Member 11rdc11's Avatar
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    Yahoo just used change of base formula.

    \log y = \frac{\ln {y}}{\ln {10}}
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