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Math Help - Evaluating a complicated integral please help urgent

  1. #1
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    Evaluating a complicated integral please help urgent

    For n any non-negative integer evaluate the integral :

     \int x^n \ln (x) dx


    Attempt to solution:

    use integration by parts

    dv=x^n
    v=\frac{x^{n-1}}{n-1}

    u=\ln (x)

    du=1/x

    \int u dv=\ln (x) \frac{x^{n-1}}{n-1}-\int\frac{x^{n-1}}{n-1} \frac{1}{x}

    I'm stuck here how do l further simplify this thing ?
    Last edited by mr fantastic; April 19th 2009 at 10:28 PM. Reason: Fixed latex tags, substituted \ln, corrected some latex
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  2. #2
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    Quote Originally Posted by nyasha View Post
    For n any non-negative integer evaluate the integral :

    \int x^n In(x) dx


    Attempt to solution:

    use integration by parts

    dv=x^n
    v=\frac{x^{n-1}}{n-1}

    u=In(x)
    du=1/x
    \int udv=In(x)\frac{x^{n-1}}{n-1}-\int\frac{x^{n-1}}{n-1}\frac{1}x}

    I'm stuck here how do l further simplify this thing ?
    \int x^n \ln{x} \, dx

    u = \ln{x} ... du = \frac{1}{x} \, dx

    dv = x^n \, dx ... v = \frac{x^{n+1}}{n+1}

    \int x^n \ln{x} \, dx = \frac{x^{n+1}}{n+1} \cdot \ln{x} - \frac{1}{n+1} \int x^n \, dx

    \int x^n \ln{x} \, dx = \frac{x^{n+1}}{n+1} \cdot \ln{x} - \frac{x^{n+1}}{(n+1)^2} + C
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  3. #3
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    Hello, nyasha!

    At this site, you must use [math ] [/math ] for LaTeX.


    For any non-negative integer n, evaluate: . \int x^n\ln(x)\,dx
    I would do it like this . . .

    . . \begin{array}{ccccccc}u &=& \ln x & & dv  &=& x^n\,dx \\ du &=& \frac{dx}{x} & & v &=& \frac{x^{n+1}}{n+1} \end{array}

    Then we have: . \frac{1}{n+1}\,x^{n+1}\ln x \;-\; \frac{1}{n+1}\int x^n\,dx \;\;=\;\;\frac{1}{n+1}\,x^{n+1}\ln x \;-\; \frac{1}{(n+1)^2}\,x^{n+1} + C

    . . . . . . = \;\frac{1}{(n+1)^2}\,x^{n+1}\bigg[(n+1)\ln x - 1\bigg] + C



    Edit: Darn, too slow again!
    .
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  4. #4
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    Quote Originally Posted by skeeter View Post
    \int x^n \ln{x} \, dx

    u = \ln{x} ... du = \frac{1}{x} \, dx

    dv = x^n \, dx ... v = \frac{x^{n+1}}{n+1}

    \int x^n \ln{x} \, dx = \frac{x^{n+1}}{n+1} \cdot \ln{x} - \frac{1}{n+1} \int x^n \, dx

    \int x^n \ln{x} \, dx = \frac{x^{n+1}}{n+1} \cdot \ln{x} - \frac{x^{n+1}}{(n+1)^2} + C

    Thanks
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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, nyasha!

    At this site, you must use [math ] [/math ] for LaTeX.


    I would do it like this . . .

    . . \begin{array}{ccccccc}u &=& \ln x & & dv  &=& x^n\,dx \\ du &=& \frac{dx}{x} & & v &=& \frac{x^{n+1}}{n+1} \end{array}

    Then we have: . \frac{1}{n+1}\,x^{n+1}\ln x \;-\; \frac{1}{n+1}\int x^n\,dx \;\;=\;\;\frac{1}{n+1}\,x^{n+1}\ln x \;-\; \frac{1}{(n+1)^2}\,x^{n+1} + C

    . . . . . . = \;\frac{1}{(n+1)^2}\,x^{n+1}\bigg[(n+1)\ln x - 1\bigg] + C



    Edit: Darn, too slow again!
    .

    Thanks,actually that was the simplified format they wanted in the answer text.
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