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Math Help - Integration and I dont get along just yet

  1. #1
    Junior Member
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    Integration and I dont get along just yet

    Could some body please help I just cant seem to get these two questions
    Integrate x^(3/2)ln x dx

    Integrate (((x^2)-x+6)/(x^3)+3x) dx

    Please try not to skip too many steps or I wont be able to follow
    Thanks in advance, Sam
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  2. #2
    Super Member wingless's Avatar
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    \int x^{\frac{3}{2}}\ln x~dx

    Let u = \ln x and dv = x^{\frac{3}{2}}~dx

    Then du = \frac{1}{x}~dx and v = \frac{2}{5}x^{\frac{5}{2}}

    \int u~dv = u\cdot v - \int v~du

    \int x^{\frac{3}{2}}\ln x~dx = \frac{2}{5}x^{\frac{5}{2}}\ln x - \int \frac{2}{5}x^{\frac{5}{2}}\frac{1}{x}~dx

    \frac{2}{5}x^{\frac{5}{2}}\ln x - \frac{2}{5}\int x^{\frac{3}{2}}~dx

    \frac{2}{5}x^{\frac{5}{2}}\ln x - \frac{2}{5}\cdot\frac{2}{5}x^{\frac{5}{2}}

    \frac{2}{25}x^{\frac{5}{2}} \cdot \left ( 5\ln x - 2\right )



    ------------------------------



    \int \frac{x^2-x+6}{x^3+3x}~dx

    \int \frac{x^2+6}{x^3+3x}~dx - \int \frac{x}{x^3+3x}~dx

    Apply partial fraction decomposition on the left integral (That will break its shell). And simplify the right integral.

    \frac{x^2+6}{x^3+3x} = \frac{2}{x}-\frac{x}{x^2+3}

    \int \frac{2}{x}-\frac{x}{x^2+3}~dx - \int \frac{1}{x^2+3}~dx

    The rest is easy,

    2\ln x - \frac{1}{2}\ln |x^2+3| - \frac{1}{\sqrt{3}}\arctan \frac{x}{\sqrt{3}}


    ------------------------------


    Notes: You can find the integral of \frac{x}{x^2+3} by substituting u=x^2+3.


    Maybe you wonder how I got that \frac{1}{\sqrt{3}}\arctan \frac{x}{\sqrt{3}}. This is how:

    \int \frac{1}{x^2+3}~dx

    \int \frac{1}{3(\frac{x^2}{3}+1)}~dx

    \int \frac{1}{3}\cdot\frac{1}{\frac{x^2}{3}+1}~dx

    \frac{1}{3}\int \frac{1}{\frac{x^2}{3}+1}~dx

    \frac{1}{3}\int \frac{1}{ { \left ( \frac{x}{\sqrt{3}}\right ) }^2+1}~dx

    Substitute u=\frac{x}{\sqrt{3}}

    Then, du = \frac{1}{\sqrt{3}}~dx

    \frac{1}{3}\int \frac{1}{u^2+1}\sqrt{3}~du

    \frac{1}{\sqrt{3}}\int \frac{1}{u^2+1}~du

    \frac{1}{\sqrt{3}}\arctan u

    \frac{1}{\sqrt{3}}\arctan \frac{x}{\sqrt{3}}
    Last edited by wingless; March 31st 2008 at 07:28 AM.
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