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Math Help - [SOLVED] Definite integral involving parts

  1. #1
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    [SOLVED] Definite integral involving parts

    \int_2^{6} \sqrt {t} ln t dt

    u = ln t
    dv = \sqrt{t}
    du = 1/t
    v = 2t^{\frac{3}{2}}
    <br />
\frac{2 ln t*t^\frac{3}{2}}{3} - \int \frac {2t^\frac{3}{2}}{3t}
    \frac{2 ln t*t^\frac{3}{2}}{3} - \frac {-2}{3} \int \frac {t^\frac{3}{2}}{t}
    \frac{2 ln t*t^\frac{3}{2}}{3} - \frac {-2}{3}* \frac {2t^\frac{3}{2}}{3}

    Could anyone point out the mistake I made? Continuing with this, my answer turns out wrong.

    EDIT: Nevermind guys, I have my answer. Sorry for the waste of topic. How do I delete my topic?
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Archduke01 View Post
    \int_2^{6} \sqrt {t} ln t dt

    u = ln t
    dv = \sqrt{t}
    du = 1/t
    v = 2t^{\frac{3}{2}}
    <br />
\frac{2 ln t*t^\frac{3}{2}}{3} - \int \frac {2t^\frac{3}{2}}{3t}
    \frac{2 ln t*t^\frac{3}{2}}{3} - \frac {-2}{3} \int \frac {t^\frac{3}{2}}{t}
    \frac{2 ln t*t^\frac{3}{2}}{3} - \frac {-2}{3}* \frac {2t^\frac{3}{2}}{3}

    Could anyone point out the mistake I made? Continuing with this, my answer turns out wrong.
    The fourth line of yours should be:
    v = \frac{2}{3}t^{\frac{3}{2}}

    and you should write du = \frac{1}{t}dt

    So,

    uv - \int v du

    = \frac{2}{3}t^{\frac{3}{2}}.ln(t) - \int \frac{2}{3}t^{\frac{3}{2}} \times \frac{1}{t} dt

    = \frac{2}{3}t^{\frac{3}{2}}.ln(t) - \frac{2}{3}\int t^{\frac{3}{2}} \times \frac{1}{t} dt

    = \frac{2}{3}t^{\frac{3}{2}}.ln(t) - \frac{2}{3}\int t^{{\frac{3}{2}}-1}  dt
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