Evaluating integral with inner derivative












2














Hi I am failing to spot how to integrate this. By parts I'm guessing? Any hints into the right direction would be helpful...





$$ int xfrac{d}{dx}bigg(ffrac{df}{dx}bigg) dx$$





where $f=f(x)$.



Would expanding the inner derivative operator work here?






integration







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2














Hi I am failing to spot how to integrate this. By parts I'm guessing? Any hints into the right direction would be helpful...





$$ int xfrac{d}{dx}bigg(ffrac{df}{dx}bigg) dx$$





where $f=f(x)$.



Would expanding the inner derivative operator work here?






integration







share|cite|improve this question






















  • Sorry, I did not see any answers at all that time. I will delete my comments. +5 from me
    – dmtri
    2 days ago












  • @Mark Viola, another proof that servers are not fast enough :)
    – dmtri
    2 days ago










  • @dmtri That has happened to me too, so no worry. Happy Holidays! ;-)
    – Mark Viola
    2 days ago














2












2








2


1





Hi I am failing to spot how to integrate this. By parts I'm guessing? Any hints into the right direction would be helpful...





$$ int xfrac{d}{dx}bigg(ffrac{df}{dx}bigg) dx$$





where $f=f(x)$.



Would expanding the inner derivative operator work here?






integration







share|cite|improve this question













Hi I am failing to spot how to integrate this. By parts I'm guessing? Any hints into the right direction would be helpful...





$$ int xfrac{d}{dx}bigg(ffrac{df}{dx}bigg) dx$$





where $f=f(x)$.



Would expanding the inner derivative operator work here?






integration




integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked 2 days ago









kroneckerdel69

257




257












  • Sorry, I did not see any answers at all that time. I will delete my comments. +5 from me
    – dmtri
    2 days ago












  • @Mark Viola, another proof that servers are not fast enough :)
    – dmtri
    2 days ago










  • @dmtri That has happened to me too, so no worry. Happy Holidays! ;-)
    – Mark Viola
    2 days ago


















  • Sorry, I did not see any answers at all that time. I will delete my comments. +5 from me
    – dmtri
    2 days ago












  • @Mark Viola, another proof that servers are not fast enough :)
    – dmtri
    2 days ago










  • @dmtri That has happened to me too, so no worry. Happy Holidays! ;-)
    – Mark Viola
    2 days ago
















Sorry, I did not see any answers at all that time. I will delete my comments. +5 from me
– dmtri
2 days ago






Sorry, I did not see any answers at all that time. I will delete my comments. +5 from me
– dmtri
2 days ago














@Mark Viola, another proof that servers are not fast enough :)
– dmtri
2 days ago




@Mark Viola, another proof that servers are not fast enough :)
– dmtri
2 days ago












@dmtri That has happened to me too, so no worry. Happy Holidays! ;-)
– Mark Viola
2 days ago




@dmtri That has happened to me too, so no worry. Happy Holidays! ;-)
– Mark Viola
2 days ago










1 Answer
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Note that $f(x)frac{df(x)}{dx}=frac12 frac{df^2(x)}{dx}$. Then, integrating by parts reveals



$$begin{align}
int xfrac{d}{dx}left(f(x)frac{df(x)}{dx}right),dx&=frac12 xfrac{df^2(x)}{dx}-frac12 int frac{df^2(x)}{dx},dx\\
&=frac12xfrac{df^2(x)}{dx}-frac12 f^2(x)+C
end{align}$$






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    1 Answer
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    1 Answer
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    Note that $f(x)frac{df(x)}{dx}=frac12 frac{df^2(x)}{dx}$. Then, integrating by parts reveals



    $$begin{align}
    int xfrac{d}{dx}left(f(x)frac{df(x)}{dx}right),dx&=frac12 xfrac{df^2(x)}{dx}-frac12 int frac{df^2(x)}{dx},dx\\
    &=frac12xfrac{df^2(x)}{dx}-frac12 f^2(x)+C
    end{align}$$






    share|cite|improve this answer


























      5














      Note that $f(x)frac{df(x)}{dx}=frac12 frac{df^2(x)}{dx}$. Then, integrating by parts reveals



      $$begin{align}
      int xfrac{d}{dx}left(f(x)frac{df(x)}{dx}right),dx&=frac12 xfrac{df^2(x)}{dx}-frac12 int frac{df^2(x)}{dx},dx\\
      &=frac12xfrac{df^2(x)}{dx}-frac12 f^2(x)+C
      end{align}$$






      share|cite|improve this answer
























        5












        5








        5






        Note that $f(x)frac{df(x)}{dx}=frac12 frac{df^2(x)}{dx}$. Then, integrating by parts reveals



        $$begin{align}
        int xfrac{d}{dx}left(f(x)frac{df(x)}{dx}right),dx&=frac12 xfrac{df^2(x)}{dx}-frac12 int frac{df^2(x)}{dx},dx\\
        &=frac12xfrac{df^2(x)}{dx}-frac12 f^2(x)+C
        end{align}$$






        share|cite|improve this answer












        Note that $f(x)frac{df(x)}{dx}=frac12 frac{df^2(x)}{dx}$. Then, integrating by parts reveals



        $$begin{align}
        int xfrac{d}{dx}left(f(x)frac{df(x)}{dx}right),dx&=frac12 xfrac{df^2(x)}{dx}-frac12 int frac{df^2(x)}{dx},dx\\
        &=frac12xfrac{df^2(x)}{dx}-frac12 f^2(x)+C
        end{align}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Mark Viola

        130k1274170




        130k1274170






























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