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    Right, I've differentiated part a) which has given me lnx + 1. However I don't know how to use this or where to start with part b), the integration bit.

    http://img96.imageshack.us/img96/9764/integrationr.png

    Could anyone help me out please?

    Much appreciated!
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    \dfrac{d}{dx}(xlnx)=lnx+1

    so, integrating both sides w.r.t. x: \displaystyle\int\frac{d}{dx}(xl  nx)\ dx=\int lnx\ dx+\int 1\ dx

    so \displaystyle\int lnx\ dx=\int\frac{d}{dx}(xlnx)\ dx-\int 1\ dx
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    (Original post by Pheylan)
    \dfrac{d}{dx}(xlnx)=lnx+1

    so, integrating both sides w.r.t. x: \displaystyle\int\frac{d}{dx}(xl  nx)\ dx=\int lnx\ dx+\int 1\ dx

    so \displaystyle\int lnx\ dx=\int\frac{d}{dx}(xlnx)\ dx-\int 1\ dx
    Sorry, I still don't really understand what is actually going on there. That's pretty much what it says in the mark scheme and I just need it explained really.
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    (Original post by themagicpiano)
    Sorry, I still don't really understand what is actually going on there. That's pretty much what it says in the mark scheme and I just need it explained really.
    well from the first part, you know that \dfrac{d}{dx}(xlnx)=lnx+1

    so if you integrate both sides of that equation (which you're allowed to do), you get \displaystyle\int\frac{d}{dx}(xl  nx)\ dx=\int lnx\ dx+\int 1\ dx

    subtract \displaystyle\int 1\ dx from both sides to get \displaystyle\int lnx\ dx on its own:

    \displaystyle\int lnx\ dx=\int\frac{d}{dx}(xlnx)\ dx-\int 1\ dx

    then you can simplify the RHS. (remember that \displaystyle\int\frac{d}{dx}(f(  x))\ dx=f(x) and \displaystyle\int x^n\ dx=\frac{x^{n+1}}{n+1}+c )
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    (Original post by Pheylan)
    well from the first part, you know that \dfrac{d}{dx}(xlnx)=lnx+1

    so if you integrate both sides of that equation (which you're allowed to do), you get \displaystyle\int\frac{d}{dx}(xl  nx)\ dx=\int lnx\ dx+\int 1\ dx

    subtract \displaystyle\int 1\ dx from both sides to get \displaystyle\int lnx\ dx on its own:

    \displaystyle\int lnx\ dx=\int\frac{d}{dx}(xlnx)\ dx-\int 1\ dx

    then you can simplify the RHS. (remember that \displaystyle\int\frac{d}{dx}(f(  x))\ dx=f(x) and \displaystyle\int x^n\ dx=\frac{x^{n+1}}{n+1}+c )
    Ah, that makes sense. Thank you for your help!
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    (Original post by Pheylan)
    well from the first part, you know that \dfrac{d}{dx}(xlnx)=lnx+1

    so if you integrate both sides of that equation (which you're allowed to do), you get \displaystyle\int\frac{d}{dx}(xl  nx)\ dx=\int lnx\ dx+\int 1\ dx

    subtract \displaystyle\int 1\ dx from both sides to get \displaystyle\int lnx\ dx on its own:

    \displaystyle\int lnx\ dx=\int\frac{d}{dx}(xlnx)\ dx-\int 1\ dx

    then you can simplify the RHS. (remember that \displaystyle\int\frac{d}{dx}(f(  x))\ dx=f(x) and \displaystyle\int x^n\ dx=\frac{x^{n+1}}{n+1}+c )
    Out of interest, since it says 'Hence, or otherwise find', which method could be used to find the integral if I didn't do part a)?

    Thanks again for your help!
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    (Original post by themagicpiano)
    Out of interest, since it says 'Hence, or otherwise find', which method could be used to find the integral if I didn't do part a)?
    have you learnt integration by parts yet? if so:

    \displaystyle\int lnx\ dx\equiv\int 1.lnx\ dx

    use IBP with u=lnx and dv/dx=1

    (Original post by themagicpiano)
    Thanks again for your help!
    no problem
 
 
 
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