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












2














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
























  • What are the restrictions on $m$ and $n$?
    – clathratus
    yesterday










  • Also is the $I_{m+1,n}$ in the last line is a typo?
    – clathratus
    yesterday
















2














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
























  • What are the restrictions on $m$ and $n$?
    – clathratus
    yesterday










  • Also is the $I_{m+1,n}$ in the last line is a typo?
    – clathratus
    yesterday














2












2








2


1





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















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 yesterday









clathratus

3,325331




3,325331










asked Jan 5 at 23:16









Anson PangAnson Pang

624




624












  • What are the restrictions on $m$ and $n$?
    – clathratus
    yesterday










  • Also is the $I_{m+1,n}$ in the last line is a typo?
    – clathratus
    yesterday


















  • What are the restrictions on $m$ and $n$?
    – clathratus
    yesterday










  • Also is the $I_{m+1,n}$ in the last line is a typo?
    – clathratus
    yesterday
















What are the restrictions on $m$ and $n$?
– clathratus
yesterday




What are the restrictions on $m$ and $n$?
– clathratus
yesterday












Also is the $I_{m+1,n}$ in the last line is a typo?
– clathratus
yesterday




Also is the $I_{m+1,n}$ in the last line is a typo?
– clathratus
yesterday










1 Answer
1






active

oldest

votes


















0














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





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063306%2fi-m-n-int-fracxmax2bxcndx-reduction-formula%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    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


























      0














      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
























        0












        0








        0






        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












        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 yesterday









        dezdichadodezdichado

        6,2551929




        6,2551929






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063306%2fi-m-n-int-fracxmax2bxcndx-reduction-formula%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Mario Kart Wii

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
            Read more

            The Binding of Isaac: Rebirth/Afterbirth

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
            Read more

            What does “Dominus providebit” mean?

            Image
            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
            Read more