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    2
    int (x)^2 ln (x) dx
    1

    = x- int 2 dx
    = x-2x

    = 2
    int [-x]
    1

    = -1? or is it 1? or is my integration wrong?

    Thanks for your time.
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    Is this what you are trying to integrate:

    x^2ln(x) dx

    between x=2 and x=1 ?
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    You seem to have started by differentiating instead of integrating. Remember  \int uv' dx = uv - \int u'v dx
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    Latex is your friend
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    (Original post by TheNack)
    Is this what you are trying to integrate:

    x^2ln(x) dx

    between x=2 and x=1 ?
    Yup it is. I just realised I didn't know how to integrate ln..I guess I got it confused and differentiated it instead. Thanks. And I really don't get that latex comment-what a waste of time and space.
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    (x^3Lnx)/3 - (x^3)/9
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    (Original post by calamityx)
    Yup it is. I just realised I didn't know how to integrate ln..I guess I got it confused and differentiated it instead. Thanks. And I really don't get that latex comment-what a waste of time and space.
    If you don't know how to integrate ln(x) then you can choose to differentiate it instead. Let:

    u=ln(x)

    and

    v'=x^2

    and then refer to Pork and Beans' post.
 
 
 
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