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    My brain literally stopped working... How would you integrate xlnx ? I'm guessing by parts?

    Can someone post a written solution step by step


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    u = lnx
    dv/dx = x

    stick it in the formula uv - \int v\frac{du}{dx}
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    one way is if you have: \displaystyle \int (ln(x))xdx =\int v \frac{du}{dx}dx= uv- \int u\frac{dv}{dx}dx

    so, use \displaystyle v=ln(x), \frac{dv}{dx}=\frac{1}{x}

    and

    \displaystyle \frac{du}{dx}=x, u=\frac{x^2}{2}

    plug in and integrate last part...

    (if you choose u and v the other way around, you have to know that the integral of ln(x) = xln(x)-x, and the parts bit gets messy)
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    so u=lnx du/dx = 1/x

    dv/dx = x dv/dx = x^2/2

    subsitute it into the formula, the last part gives me: intergrate x/2... is that just xln2 if you intergrate x/2?
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    (Original post by Gary)
    intergrate x/2... is that just xln2 if you intergrate x/2?
    No

    Hint:

    \dfrac{x}{2} \equiv \dfrac{1}{2}x

    And I'm sure you can integrate that.
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    I'm sure this is C4?
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    (Original post by usycool1)
    No

    Hint:

    \dfrac{x}{2} \equiv \dfrac{1}{2}x

    And I'm sure you can integrate that.
    Oh yeh of course -.- I hate myself at times ....


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    (Original post by nm786)
    I'm sure this is C4?
    It's C4 for Edexcel but other exam boards (like AQA, IIRC) have integration in C3.
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    you make lnx = u and make x= dv/dx, natural logs take priority over polynomials never forget that. Dont forget to integrate ln(x) on its own you do u=lnx and dv/dx= 1
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    (Original post by usycool1)
    It's C4 for Edexcel but other exam boards (like AQA, IIRC) have integration in C3.
    alright cool.
 
 
 
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