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Math Help - Definite Integrals for Integration by Parts

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    Definite Integrals for Integration by Parts

    I'm having a bit of difficulty with this one. Can anyone help?
    Many thanks.

    Q. \int^2_1 \frac{1}{x^2} logex dx

    Attempt: u = logex => du = \frac{1}{x} dx
    dv = \frac{1}{x^2} => dv = x-2 => v = \intx-2 => \frac{-x^-^2}{2}
    \intu dv = uv - \intv du => \int \frac{1}{x^2}logex dx => logex (\frac{-x^2}{2}) + \int \frac{x^-^2}{2} (\frac{1}{x}) => \frac{-x^-^2log_ex}{2} + \int \frac{x^-^3}{2} => \frac{-x^-^2log_ex}{2} - \frac{x^-^2}{4}
    \int^2_1 \frac{1}{x^2}logex dx = [\frac{-x^-^2log_ex}{2} - \frac{x^-^2}{4}]^2_1 => [\frac{-2^-^2log_e2}{2} - \frac{2^-^2}{4}] - [\frac{-1^-^2log_e1}{2} - \frac{1^-^2}{4} => \frac{-log_e2}{8} - \frac{1}{16} + 0 + \frac{1}{4} => \frac{-log_e2}{8} - \frac{5}{16}

    Ans.:
    (From text book): \frac{1}{2}(1 - ln2)
    Last edited by GrigOrig99; April 19th 2012 at 01:42 PM.
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    Re: Definite Integrals for Integration by Parts

    Check \int x^{-2}dx
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    Re: Definite Integrals for Integration by Parts

    I've made the change I think you're referring to but I'm still off the mark...

    u = logex => du = \frac{1}{x} dx
    dv = \frac{1}{x^2} => => v = \int \frac{1}{x^2} => ln(x2)
    \intu dv = uv - \intv du => \int \frac{1}{x^2}logex dx => logex(ln(x2)) - \intln(x2) (\frac{1}{x}) => ln(x2)logex - (x2ln(x2) - x2) (\frac{1}{x})
    \int^2_1 \frac{1}{x^2}logex dx = [ln(x2)logex - (x2ln(x2) - x2) (\frac{1}{x})] ^2_1 => [ln(4)loge2 - (4ln(4) - 4) (\frac{1}{2})] - ln(1)loge1 - (1ln(1) - 1)(1)] => ln(4)loge2 - (4ln(4) - 4) (\frac{1}{2}) + 0 + 1

    Ans.:
    (From text book): \frac{1}{2}(1 - ln2)
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    Re: Definite Integrals for Integration by Parts

    Quote Originally Posted by GrigOrig99 View Post
    I've made the change I think you're referring to but I'm still off the mark...

    u = logex => du = \frac{1}{x} dx
    dv = \frac{1}{x^2} => => v = \int \frac{1}{x^2} => ln(x2) no
    \intu dv = uv - \intv du => \int \frac{1}{x^2}logex dx => logex(ln(x2)) - \intln(x2) (\frac{1}{x}) => ln(x2)logex - (x2ln(x2) - x2) (\frac{1}{x})
    \int^2_1 \frac{1}{x^2}logex dx = [ln(x2)logex - (x2ln(x2) - x2) (\frac{1}{x})] ^2_1 => [ln(4)loge2 - (4ln(4) - 4) (\frac{1}{2})] - ln(1)loge1 - (1ln(1) - 1)(1)] => ln(4)loge2 - (4ln(4) - 4) (\frac{1}{2}) + 0 + 1

    Ans.:
    (From text book): \frac{1}{2}(1 - ln2)
    \int \frac{1}{x^2} \, dx = -\frac{1}{x} + C
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    Re: Definite Integrals for Integration by Parts

    Quote Originally Posted by GrigOrig99 View Post
    Ans.: [/B](From text book): \frac{1}{2}(1 - ln2)
    Take a look at this page.

    And this.
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    Re: Definite Integrals for Integration by Parts

    Thanks for that link. Really useful! The only question I have left is with the expression of the answer. The book has it as 1/2(1 - ln2) but the link you provided shows a final answer of 1/2(1 - log2). Is there no difference?
    Thanks again, guys.
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    Re: Definite Integrals for Integration by Parts

    Quote Originally Posted by GrigOrig99 View Post
    Thanks for that link. Really useful! The only question I have left is with the expression of the answer. The book has it as 1/2(1 - ln2) but the link you provided shows a final answer of 1/2(1 - log2). Is there no difference?
    Thanks again, guys.
    none ... in "big boy" math, log is understood base e
    Thanks from Plato
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