$I_{m,n}=intfrac{x^m}{(ax^2+bx+c)^n}dx$ Reduction Formula












2












$begingroup$


I'm having trouble proving the following reduction formula:



If



$$I_{m,n}=intfrac{x^m}{(ax^2+bx+c)^n}dx$$



then



$$intfrac{x^m}{(ax^2+bx+c)^n}dx=-frac{x^{m-1}}{a(2m-n-1)(ax^2+bx+c)^{n-1}}-frac{b(n-m)}{a(2m-n-1)}I_{m-1,n}+frac{c(m-1)}{a(2m-n-1)}I_{m-2,n}$$



My attempt went as follows:



$$intfrac{x^m}{(ax^2+bx+c)^n}dxspacebegin{vmatrix}u=frac{1}{(ax^2+bx+c)^n}\du=frac{-n(2ax+b)}{(ax^2+bx+c)^{n+1}}dxend{vmatrix}space dv=x^mdxquad v=frac{1}{m+1}x^{m+1}\intfrac{x^m}{(ax^2+bx+c)^n}dx=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{n}{m+1}bigg)intfrac{(2ax+b)x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{n}{m+1}bigg)bigg(2aintfrac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\+bintfrac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dxbigg)\=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{2an}{m+1}bigg)intfrac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\+bigg(frac{bn}{m+1}bigg)intfrac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\=frac{1}{m+1}I_{m+1,n}+bigg(frac{2an}{m+1}bigg)I_{m+2,n+1}+bigg(frac{bn}{m+1}bigg)I_{m+1,n+1}$$



This is where I'm stuck, any help?






integration reduction-formula







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  • $begingroup$
    What are the restrictions on $m$ and $n$?
    $endgroup$
    – clathratus
    Jan 8 at 2:49










  • $begingroup$
    Also is the $I_{m+1,n}$ in the last line is a typo?
    $endgroup$
    – clathratus
    Jan 8 at 2:50
















2












$begingroup$


I'm having trouble proving the following reduction formula:



If



$$I_{m,n}=intfrac{x^m}{(ax^2+bx+c)^n}dx$$



then



$$intfrac{x^m}{(ax^2+bx+c)^n}dx=-frac{x^{m-1}}{a(2m-n-1)(ax^2+bx+c)^{n-1}}-frac{b(n-m)}{a(2m-n-1)}I_{m-1,n}+frac{c(m-1)}{a(2m-n-1)}I_{m-2,n}$$



My attempt went as follows:



$$intfrac{x^m}{(ax^2+bx+c)^n}dxspacebegin{vmatrix}u=frac{1}{(ax^2+bx+c)^n}\du=frac{-n(2ax+b)}{(ax^2+bx+c)^{n+1}}dxend{vmatrix}space dv=x^mdxquad v=frac{1}{m+1}x^{m+1}\intfrac{x^m}{(ax^2+bx+c)^n}dx=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{n}{m+1}bigg)intfrac{(2ax+b)x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{n}{m+1}bigg)bigg(2aintfrac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\+bintfrac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dxbigg)\=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{2an}{m+1}bigg)intfrac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\+bigg(frac{bn}{m+1}bigg)intfrac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\=frac{1}{m+1}I_{m+1,n}+bigg(frac{2an}{m+1}bigg)I_{m+2,n+1}+bigg(frac{bn}{m+1}bigg)I_{m+1,n+1}$$



This is where I'm stuck, any help?






integration reduction-formula







share|cite|improve this question











$endgroup$












  • $begingroup$
    What are the restrictions on $m$ and $n$?
    $endgroup$
    – clathratus
    Jan 8 at 2:49










  • $begingroup$
    Also is the $I_{m+1,n}$ in the last line is a typo?
    $endgroup$
    – clathratus
    Jan 8 at 2:50














2












2








2


1



$begingroup$


I'm having trouble proving the following reduction formula:



If



$$I_{m,n}=intfrac{x^m}{(ax^2+bx+c)^n}dx$$



then



$$intfrac{x^m}{(ax^2+bx+c)^n}dx=-frac{x^{m-1}}{a(2m-n-1)(ax^2+bx+c)^{n-1}}-frac{b(n-m)}{a(2m-n-1)}I_{m-1,n}+frac{c(m-1)}{a(2m-n-1)}I_{m-2,n}$$



My attempt went as follows:



$$intfrac{x^m}{(ax^2+bx+c)^n}dxspacebegin{vmatrix}u=frac{1}{(ax^2+bx+c)^n}\du=frac{-n(2ax+b)}{(ax^2+bx+c)^{n+1}}dxend{vmatrix}space dv=x^mdxquad v=frac{1}{m+1}x^{m+1}\intfrac{x^m}{(ax^2+bx+c)^n}dx=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{n}{m+1}bigg)intfrac{(2ax+b)x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{n}{m+1}bigg)bigg(2aintfrac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\+bintfrac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dxbigg)\=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{2an}{m+1}bigg)intfrac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\+bigg(frac{bn}{m+1}bigg)intfrac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\=frac{1}{m+1}I_{m+1,n}+bigg(frac{2an}{m+1}bigg)I_{m+2,n+1}+bigg(frac{bn}{m+1}bigg)I_{m+1,n+1}$$



This is where I'm stuck, any help?






integration reduction-formula







share|cite|improve this question











$endgroup$




I'm having trouble proving the following reduction formula:



If



$$I_{m,n}=intfrac{x^m}{(ax^2+bx+c)^n}dx$$



then



$$intfrac{x^m}{(ax^2+bx+c)^n}dx=-frac{x^{m-1}}{a(2m-n-1)(ax^2+bx+c)^{n-1}}-frac{b(n-m)}{a(2m-n-1)}I_{m-1,n}+frac{c(m-1)}{a(2m-n-1)}I_{m-2,n}$$



My attempt went as follows:



$$intfrac{x^m}{(ax^2+bx+c)^n}dxspacebegin{vmatrix}u=frac{1}{(ax^2+bx+c)^n}\du=frac{-n(2ax+b)}{(ax^2+bx+c)^{n+1}}dxend{vmatrix}space dv=x^mdxquad v=frac{1}{m+1}x^{m+1}\intfrac{x^m}{(ax^2+bx+c)^n}dx=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{n}{m+1}bigg)intfrac{(2ax+b)x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{n}{m+1}bigg)bigg(2aintfrac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\+bintfrac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dxbigg)\=frac{x^{m+1}}{(m+1)(ax^2+bx+c)^n}+bigg(frac{2an}{m+1}bigg)intfrac{x^{m+2}}{(ax^2+bx+c)^{n+1}}dx\+bigg(frac{bn}{m+1}bigg)intfrac{x^{m+1}}{(ax^2+bx+c)^{n+1}}dx\=frac{1}{m+1}I_{m+1,n}+bigg(frac{2an}{m+1}bigg)I_{m+2,n+1}+bigg(frac{bn}{m+1}bigg)I_{m+1,n+1}$$



This is where I'm stuck, any help?






integration reduction-formula




integration reduction-formula






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 2:37









clathratus

4,412336




4,412336










asked Jan 5 at 23:16









Anson PangAnson Pang

6215




6215












  • $begingroup$
    What are the restrictions on $m$ and $n$?
    $endgroup$
    – clathratus
    Jan 8 at 2:49










  • $begingroup$
    Also is the $I_{m+1,n}$ in the last line is a typo?
    $endgroup$
    – clathratus
    Jan 8 at 2:50


















  • $begingroup$
    What are the restrictions on $m$ and $n$?
    $endgroup$
    – clathratus
    Jan 8 at 2:49










  • $begingroup$
    Also is the $I_{m+1,n}$ in the last line is a typo?
    $endgroup$
    – clathratus
    Jan 8 at 2:50
















$begingroup$
What are the restrictions on $m$ and $n$?
$endgroup$
– clathratus
Jan 8 at 2:49




$begingroup$
What are the restrictions on $m$ and $n$?
$endgroup$
– clathratus
Jan 8 at 2:49












$begingroup$
Also is the $I_{m+1,n}$ in the last line is a typo?
$endgroup$
– clathratus
Jan 8 at 2:50




$begingroup$
Also is the $I_{m+1,n}$ in the last line is a typo?
$endgroup$
– clathratus
Jan 8 at 2:50










1 Answer
1






active

oldest

votes


















0












$begingroup$

So this means you are using the integration by parts in the wrong direction or with a wrong choice. In the reduction formula you are trying to achieve, you can notice that $m$ is decreasing while $n$ stays constant. Thus, you need a choice of $u,v$ such that:
$$udv = dfrac{x^mdx}{f^n(x)},$$
where $f(x) = ax^2+bx+c$ and:
$$vdu = Cdfrac{x^{m-1}dx}{f^n(x)},$$
with $C$ some constant. If you tried to naively take $u = x^m,$ then you will have $v = intdfrac{dx}{f^n(x)},$ which you cannot evaluate. Therefore, one needs to improvise somehow. For instance,
$$dleft(dfrac{1}{g^k(x)}right) = dfrac{kg'(x)g^{k-1}(x)dx}{g^{2k}(x)} = kdfrac{g'(x)dx}{g^{k+1}(x)}$$ for appropriately smooth function $g(x).$ This is where I will leave you with a hint:
$$dfrac{x^mdx}{f^n(x)} = dfrac{x^{m-1}}{2a}cdotdfrac{d(ax^2+bx+c)}{(ax^2+bx+c)^n} -dfrac{b}{2a}cdotdfrac{x^{m-1}}{(ax^2+bx+c)^n}=dots $$






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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    So this means you are using the integration by parts in the wrong direction or with a wrong choice. In the reduction formula you are trying to achieve, you can notice that $m$ is decreasing while $n$ stays constant. Thus, you need a choice of $u,v$ such that:
    $$udv = dfrac{x^mdx}{f^n(x)},$$
    where $f(x) = ax^2+bx+c$ and:
    $$vdu = Cdfrac{x^{m-1}dx}{f^n(x)},$$
    with $C$ some constant. If you tried to naively take $u = x^m,$ then you will have $v = intdfrac{dx}{f^n(x)},$ which you cannot evaluate. Therefore, one needs to improvise somehow. For instance,
    $$dleft(dfrac{1}{g^k(x)}right) = dfrac{kg'(x)g^{k-1}(x)dx}{g^{2k}(x)} = kdfrac{g'(x)dx}{g^{k+1}(x)}$$ for appropriately smooth function $g(x).$ This is where I will leave you with a hint:
    $$dfrac{x^mdx}{f^n(x)} = dfrac{x^{m-1}}{2a}cdotdfrac{d(ax^2+bx+c)}{(ax^2+bx+c)^n} -dfrac{b}{2a}cdotdfrac{x^{m-1}}{(ax^2+bx+c)^n}=dots $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      So this means you are using the integration by parts in the wrong direction or with a wrong choice. In the reduction formula you are trying to achieve, you can notice that $m$ is decreasing while $n$ stays constant. Thus, you need a choice of $u,v$ such that:
      $$udv = dfrac{x^mdx}{f^n(x)},$$
      where $f(x) = ax^2+bx+c$ and:
      $$vdu = Cdfrac{x^{m-1}dx}{f^n(x)},$$
      with $C$ some constant. If you tried to naively take $u = x^m,$ then you will have $v = intdfrac{dx}{f^n(x)},$ which you cannot evaluate. Therefore, one needs to improvise somehow. For instance,
      $$dleft(dfrac{1}{g^k(x)}right) = dfrac{kg'(x)g^{k-1}(x)dx}{g^{2k}(x)} = kdfrac{g'(x)dx}{g^{k+1}(x)}$$ for appropriately smooth function $g(x).$ This is where I will leave you with a hint:
      $$dfrac{x^mdx}{f^n(x)} = dfrac{x^{m-1}}{2a}cdotdfrac{d(ax^2+bx+c)}{(ax^2+bx+c)^n} -dfrac{b}{2a}cdotdfrac{x^{m-1}}{(ax^2+bx+c)^n}=dots $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        So this means you are using the integration by parts in the wrong direction or with a wrong choice. In the reduction formula you are trying to achieve, you can notice that $m$ is decreasing while $n$ stays constant. Thus, you need a choice of $u,v$ such that:
        $$udv = dfrac{x^mdx}{f^n(x)},$$
        where $f(x) = ax^2+bx+c$ and:
        $$vdu = Cdfrac{x^{m-1}dx}{f^n(x)},$$
        with $C$ some constant. If you tried to naively take $u = x^m,$ then you will have $v = intdfrac{dx}{f^n(x)},$ which you cannot evaluate. Therefore, one needs to improvise somehow. For instance,
        $$dleft(dfrac{1}{g^k(x)}right) = dfrac{kg'(x)g^{k-1}(x)dx}{g^{2k}(x)} = kdfrac{g'(x)dx}{g^{k+1}(x)}$$ for appropriately smooth function $g(x).$ This is where I will leave you with a hint:
        $$dfrac{x^mdx}{f^n(x)} = dfrac{x^{m-1}}{2a}cdotdfrac{d(ax^2+bx+c)}{(ax^2+bx+c)^n} -dfrac{b}{2a}cdotdfrac{x^{m-1}}{(ax^2+bx+c)^n}=dots $$






        share|cite|improve this answer









        $endgroup$



        So this means you are using the integration by parts in the wrong direction or with a wrong choice. In the reduction formula you are trying to achieve, you can notice that $m$ is decreasing while $n$ stays constant. Thus, you need a choice of $u,v$ such that:
        $$udv = dfrac{x^mdx}{f^n(x)},$$
        where $f(x) = ax^2+bx+c$ and:
        $$vdu = Cdfrac{x^{m-1}dx}{f^n(x)},$$
        with $C$ some constant. If you tried to naively take $u = x^m,$ then you will have $v = intdfrac{dx}{f^n(x)},$ which you cannot evaluate. Therefore, one needs to improvise somehow. For instance,
        $$dleft(dfrac{1}{g^k(x)}right) = dfrac{kg'(x)g^{k-1}(x)dx}{g^{2k}(x)} = kdfrac{g'(x)dx}{g^{k+1}(x)}$$ for appropriately smooth function $g(x).$ This is where I will leave you with a hint:
        $$dfrac{x^mdx}{f^n(x)} = dfrac{x^{m-1}}{2a}cdotdfrac{d(ax^2+bx+c)}{(ax^2+bx+c)^n} -dfrac{b}{2a}cdotdfrac{x^{m-1}}{(ax^2+bx+c)^n}=dots $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 2:59









        dezdichadodezdichado

        6,3711929




        6,3711929






























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