Integral check. Is partial fractions the only way?












1












$begingroup$


I'm getting a different answer from wolfram and I have no idea where. I have to integrate:



$$int_0^1 frac{xdx}{(2x+1)^3}$$



Is partial fractions the only way?



So evaluating the fraction first:



$$frac{x}{(2x+1)^3} = frac{A}{2x+1} + frac{B}{(2x+1)^2} + frac{C}{(2x+1)^3}$$



$$x = A(2x+1)^2 + B(2x+1) + C$$



$$x = A(4x^2 + 4x + 1) + 2Bx + B + C$$



$$x = 4Ax^2 + 4Ax + A + 2Bx + B + C$$



$$x = x^2(4A) + x(4A+2B) + A + B + C$$



$4A = 0$ and $4A+2B = 1$ and $A + B + C = 0$ so $A = 0$ and $B=frac{1}{2}$ and $C = frac{-1}{2}$



Is the partial fraction part right?



So then I get:



$$int_0^1 frac{xdx}{(2x+1)^3} = int_0^1 frac{dx}{(2(2x+1)^2)} - int_0^1 frac{dx}{2(2x+1)^3}$$



for both, using $u = 2x+1$ and $frac{du}{dx} = 2$ and $du = 2dx$ and $frac{du}{2} = dx$,
$$frac{1}{4} int frac{du}{u^2} - frac{1}{4} int frac{du}{u^3}$$



$$ = [frac{1}{4} - u^{-1} - frac{1}{4} cdot frac{1}{-2} u^{-2} ]_1^3$$



I'm going the route of changing the limits to the new u and since $u = 2x+1$, when $x = 0, u = 1$ and when $x = 1, u = 3$. Is this the right path?



finally I get



$$[frac{-1}{4}u^{-1} + frac{1}{8}u^{-2} ]_1^3$$



I plug in numbers but I get a different answer than wolfram...






integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    I can't see any mistake in your work. What's the answer given by wolfram?
    $endgroup$
    – Thomas Shelby
    2 hours ago










  • $begingroup$
    They have 1/18 as the answer @ThomasShelby
    $endgroup$
    – Jwan622
    38 mins ago
















1












$begingroup$


I'm getting a different answer from wolfram and I have no idea where. I have to integrate:



$$int_0^1 frac{xdx}{(2x+1)^3}$$



Is partial fractions the only way?



So evaluating the fraction first:



$$frac{x}{(2x+1)^3} = frac{A}{2x+1} + frac{B}{(2x+1)^2} + frac{C}{(2x+1)^3}$$



$$x = A(2x+1)^2 + B(2x+1) + C$$



$$x = A(4x^2 + 4x + 1) + 2Bx + B + C$$



$$x = 4Ax^2 + 4Ax + A + 2Bx + B + C$$



$$x = x^2(4A) + x(4A+2B) + A + B + C$$



$4A = 0$ and $4A+2B = 1$ and $A + B + C = 0$ so $A = 0$ and $B=frac{1}{2}$ and $C = frac{-1}{2}$



Is the partial fraction part right?



So then I get:



$$int_0^1 frac{xdx}{(2x+1)^3} = int_0^1 frac{dx}{(2(2x+1)^2)} - int_0^1 frac{dx}{2(2x+1)^3}$$



for both, using $u = 2x+1$ and $frac{du}{dx} = 2$ and $du = 2dx$ and $frac{du}{2} = dx$,
$$frac{1}{4} int frac{du}{u^2} - frac{1}{4} int frac{du}{u^3}$$



$$ = [frac{1}{4} - u^{-1} - frac{1}{4} cdot frac{1}{-2} u^{-2} ]_1^3$$



I'm going the route of changing the limits to the new u and since $u = 2x+1$, when $x = 0, u = 1$ and when $x = 1, u = 3$. Is this the right path?



finally I get



$$[frac{-1}{4}u^{-1} + frac{1}{8}u^{-2} ]_1^3$$



I plug in numbers but I get a different answer than wolfram...






integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    I can't see any mistake in your work. What's the answer given by wolfram?
    $endgroup$
    – Thomas Shelby
    2 hours ago










  • $begingroup$
    They have 1/18 as the answer @ThomasShelby
    $endgroup$
    – Jwan622
    38 mins ago














1












1








1





$begingroup$


I'm getting a different answer from wolfram and I have no idea where. I have to integrate:



$$int_0^1 frac{xdx}{(2x+1)^3}$$



Is partial fractions the only way?



So evaluating the fraction first:



$$frac{x}{(2x+1)^3} = frac{A}{2x+1} + frac{B}{(2x+1)^2} + frac{C}{(2x+1)^3}$$



$$x = A(2x+1)^2 + B(2x+1) + C$$



$$x = A(4x^2 + 4x + 1) + 2Bx + B + C$$



$$x = 4Ax^2 + 4Ax + A + 2Bx + B + C$$



$$x = x^2(4A) + x(4A+2B) + A + B + C$$



$4A = 0$ and $4A+2B = 1$ and $A + B + C = 0$ so $A = 0$ and $B=frac{1}{2}$ and $C = frac{-1}{2}$



Is the partial fraction part right?



So then I get:



$$int_0^1 frac{xdx}{(2x+1)^3} = int_0^1 frac{dx}{(2(2x+1)^2)} - int_0^1 frac{dx}{2(2x+1)^3}$$



for both, using $u = 2x+1$ and $frac{du}{dx} = 2$ and $du = 2dx$ and $frac{du}{2} = dx$,
$$frac{1}{4} int frac{du}{u^2} - frac{1}{4} int frac{du}{u^3}$$



$$ = [frac{1}{4} - u^{-1} - frac{1}{4} cdot frac{1}{-2} u^{-2} ]_1^3$$



I'm going the route of changing the limits to the new u and since $u = 2x+1$, when $x = 0, u = 1$ and when $x = 1, u = 3$. Is this the right path?



finally I get



$$[frac{-1}{4}u^{-1} + frac{1}{8}u^{-2} ]_1^3$$



I plug in numbers but I get a different answer than wolfram...






integration







share|cite|improve this question











$endgroup$




I'm getting a different answer from wolfram and I have no idea where. I have to integrate:



$$int_0^1 frac{xdx}{(2x+1)^3}$$



Is partial fractions the only way?



So evaluating the fraction first:



$$frac{x}{(2x+1)^3} = frac{A}{2x+1} + frac{B}{(2x+1)^2} + frac{C}{(2x+1)^3}$$



$$x = A(2x+1)^2 + B(2x+1) + C$$



$$x = A(4x^2 + 4x + 1) + 2Bx + B + C$$



$$x = 4Ax^2 + 4Ax + A + 2Bx + B + C$$



$$x = x^2(4A) + x(4A+2B) + A + B + C$$



$4A = 0$ and $4A+2B = 1$ and $A + B + C = 0$ so $A = 0$ and $B=frac{1}{2}$ and $C = frac{-1}{2}$



Is the partial fraction part right?



So then I get:



$$int_0^1 frac{xdx}{(2x+1)^3} = int_0^1 frac{dx}{(2(2x+1)^2)} - int_0^1 frac{dx}{2(2x+1)^3}$$



for both, using $u = 2x+1$ and $frac{du}{dx} = 2$ and $du = 2dx$ and $frac{du}{2} = dx$,
$$frac{1}{4} int frac{du}{u^2} - frac{1}{4} int frac{du}{u^3}$$



$$ = [frac{1}{4} - u^{-1} - frac{1}{4} cdot frac{1}{-2} u^{-2} ]_1^3$$



I'm going the route of changing the limits to the new u and since $u = 2x+1$, when $x = 0, u = 1$ and when $x = 1, u = 3$. Is this the right path?



finally I get



$$[frac{-1}{4}u^{-1} + frac{1}{8}u^{-2} ]_1^3$$



I plug in numbers but I get a different answer than wolfram...






integration




integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Joshua Mundinger

2,5191028




2,5191028










asked 2 hours ago









Jwan622Jwan622

2,13611530




2,13611530












  • $begingroup$
    I can't see any mistake in your work. What's the answer given by wolfram?
    $endgroup$
    – Thomas Shelby
    2 hours ago










  • $begingroup$
    They have 1/18 as the answer @ThomasShelby
    $endgroup$
    – Jwan622
    38 mins ago


















  • $begingroup$
    I can't see any mistake in your work. What's the answer given by wolfram?
    $endgroup$
    – Thomas Shelby
    2 hours ago










  • $begingroup$
    They have 1/18 as the answer @ThomasShelby
    $endgroup$
    – Jwan622
    38 mins ago
















$begingroup$
I can't see any mistake in your work. What's the answer given by wolfram?
$endgroup$
– Thomas Shelby
2 hours ago




$begingroup$
I can't see any mistake in your work. What's the answer given by wolfram?
$endgroup$
– Thomas Shelby
2 hours ago












$begingroup$
They have 1/18 as the answer @ThomasShelby
$endgroup$
– Jwan622
38 mins ago




$begingroup$
They have 1/18 as the answer @ThomasShelby
$endgroup$
– Jwan622
38 mins ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

Hint: Substitute $x=tfrac12 u -tfrac12$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You beat me to it.
    $endgroup$
    – randomgirl
    2 hours ago










  • $begingroup$
    If I go your way, I wind up with $frac{1}{2} int frac{1}{u^2} - frac{1}{2} int frac{1}{u^3}$ which is different than what I have at some point. I can't see my error.
    $endgroup$
    – Jwan622
    16 mins ago



















2












$begingroup$

A much much easier way to solve it is by using integration by parts.



Hint: Take $u=x$, $dv = frac{dx}{(2x+1)^3}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $u$ but then $dv$?
    $endgroup$
    – manooooh
    2 hours ago










  • $begingroup$
    @manooooh What do you mean? I don't understand.
    $endgroup$
    – Haris Gusic
    2 hours ago










  • $begingroup$
    I am sorry, I understood that you used sub. My apologies.
    $endgroup$
    – manooooh
    2 hours ago



















1












$begingroup$

1) Observe: $$x= frac 12 times left ( (2x + 1) -1right).$$



2) Use this to obtain $$frac{x}{(2x+1)^3} = frac{1}{2} times frac{1}{(2x+1)^2} - frac 12 times frac{1}{(2x+1)^3}.$$



3) Integrate to get



$$ int frac{x}{(2x+1)^3} dx = -frac{1}{4} (2x+1)^{-1} + frac{1}{8} (2x+1)^{-2}.$$






share|cite|improve this answer









$endgroup$













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Hint: Substitute $x=tfrac12 u -tfrac12$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      You beat me to it.
      $endgroup$
      – randomgirl
      2 hours ago










    • $begingroup$
      If I go your way, I wind up with $frac{1}{2} int frac{1}{u^2} - frac{1}{2} int frac{1}{u^3}$ which is different than what I have at some point. I can't see my error.
      $endgroup$
      – Jwan622
      16 mins ago
















    3












    $begingroup$

    Hint: Substitute $x=tfrac12 u -tfrac12$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      You beat me to it.
      $endgroup$
      – randomgirl
      2 hours ago










    • $begingroup$
      If I go your way, I wind up with $frac{1}{2} int frac{1}{u^2} - frac{1}{2} int frac{1}{u^3}$ which is different than what I have at some point. I can't see my error.
      $endgroup$
      – Jwan622
      16 mins ago














    3












    3








    3





    $begingroup$

    Hint: Substitute $x=tfrac12 u -tfrac12$






    share|cite|improve this answer









    $endgroup$



    Hint: Substitute $x=tfrac12 u -tfrac12$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 hours ago









    MPWMPW

    30.3k12157




    30.3k12157












    • $begingroup$
      You beat me to it.
      $endgroup$
      – randomgirl
      2 hours ago










    • $begingroup$
      If I go your way, I wind up with $frac{1}{2} int frac{1}{u^2} - frac{1}{2} int frac{1}{u^3}$ which is different than what I have at some point. I can't see my error.
      $endgroup$
      – Jwan622
      16 mins ago


















    • $begingroup$
      You beat me to it.
      $endgroup$
      – randomgirl
      2 hours ago










    • $begingroup$
      If I go your way, I wind up with $frac{1}{2} int frac{1}{u^2} - frac{1}{2} int frac{1}{u^3}$ which is different than what I have at some point. I can't see my error.
      $endgroup$
      – Jwan622
      16 mins ago
















    $begingroup$
    You beat me to it.
    $endgroup$
    – randomgirl
    2 hours ago




    $begingroup$
    You beat me to it.
    $endgroup$
    – randomgirl
    2 hours ago












    $begingroup$
    If I go your way, I wind up with $frac{1}{2} int frac{1}{u^2} - frac{1}{2} int frac{1}{u^3}$ which is different than what I have at some point. I can't see my error.
    $endgroup$
    – Jwan622
    16 mins ago




    $begingroup$
    If I go your way, I wind up with $frac{1}{2} int frac{1}{u^2} - frac{1}{2} int frac{1}{u^3}$ which is different than what I have at some point. I can't see my error.
    $endgroup$
    – Jwan622
    16 mins ago











    2












    $begingroup$

    A much much easier way to solve it is by using integration by parts.



    Hint: Take $u=x$, $dv = frac{dx}{(2x+1)^3}$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      $u$ but then $dv$?
      $endgroup$
      – manooooh
      2 hours ago










    • $begingroup$
      @manooooh What do you mean? I don't understand.
      $endgroup$
      – Haris Gusic
      2 hours ago










    • $begingroup$
      I am sorry, I understood that you used sub. My apologies.
      $endgroup$
      – manooooh
      2 hours ago
















    2












    $begingroup$

    A much much easier way to solve it is by using integration by parts.



    Hint: Take $u=x$, $dv = frac{dx}{(2x+1)^3}$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      $u$ but then $dv$?
      $endgroup$
      – manooooh
      2 hours ago










    • $begingroup$
      @manooooh What do you mean? I don't understand.
      $endgroup$
      – Haris Gusic
      2 hours ago










    • $begingroup$
      I am sorry, I understood that you used sub. My apologies.
      $endgroup$
      – manooooh
      2 hours ago














    2












    2








    2





    $begingroup$

    A much much easier way to solve it is by using integration by parts.



    Hint: Take $u=x$, $dv = frac{dx}{(2x+1)^3}$.






    share|cite|improve this answer









    $endgroup$



    A much much easier way to solve it is by using integration by parts.



    Hint: Take $u=x$, $dv = frac{dx}{(2x+1)^3}$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 hours ago









    Haris GusicHaris Gusic

    960116




    960116












    • $begingroup$
      $u$ but then $dv$?
      $endgroup$
      – manooooh
      2 hours ago










    • $begingroup$
      @manooooh What do you mean? I don't understand.
      $endgroup$
      – Haris Gusic
      2 hours ago










    • $begingroup$
      I am sorry, I understood that you used sub. My apologies.
      $endgroup$
      – manooooh
      2 hours ago


















    • $begingroup$
      $u$ but then $dv$?
      $endgroup$
      – manooooh
      2 hours ago










    • $begingroup$
      @manooooh What do you mean? I don't understand.
      $endgroup$
      – Haris Gusic
      2 hours ago










    • $begingroup$
      I am sorry, I understood that you used sub. My apologies.
      $endgroup$
      – manooooh
      2 hours ago
















    $begingroup$
    $u$ but then $dv$?
    $endgroup$
    – manooooh
    2 hours ago




    $begingroup$
    $u$ but then $dv$?
    $endgroup$
    – manooooh
    2 hours ago












    $begingroup$
    @manooooh What do you mean? I don't understand.
    $endgroup$
    – Haris Gusic
    2 hours ago




    $begingroup$
    @manooooh What do you mean? I don't understand.
    $endgroup$
    – Haris Gusic
    2 hours ago












    $begingroup$
    I am sorry, I understood that you used sub. My apologies.
    $endgroup$
    – manooooh
    2 hours ago




    $begingroup$
    I am sorry, I understood that you used sub. My apologies.
    $endgroup$
    – manooooh
    2 hours ago











    1












    $begingroup$

    1) Observe: $$x= frac 12 times left ( (2x + 1) -1right).$$



    2) Use this to obtain $$frac{x}{(2x+1)^3} = frac{1}{2} times frac{1}{(2x+1)^2} - frac 12 times frac{1}{(2x+1)^3}.$$



    3) Integrate to get



    $$ int frac{x}{(2x+1)^3} dx = -frac{1}{4} (2x+1)^{-1} + frac{1}{8} (2x+1)^{-2}.$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      1) Observe: $$x= frac 12 times left ( (2x + 1) -1right).$$



      2) Use this to obtain $$frac{x}{(2x+1)^3} = frac{1}{2} times frac{1}{(2x+1)^2} - frac 12 times frac{1}{(2x+1)^3}.$$



      3) Integrate to get



      $$ int frac{x}{(2x+1)^3} dx = -frac{1}{4} (2x+1)^{-1} + frac{1}{8} (2x+1)^{-2}.$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        1) Observe: $$x= frac 12 times left ( (2x + 1) -1right).$$



        2) Use this to obtain $$frac{x}{(2x+1)^3} = frac{1}{2} times frac{1}{(2x+1)^2} - frac 12 times frac{1}{(2x+1)^3}.$$



        3) Integrate to get



        $$ int frac{x}{(2x+1)^3} dx = -frac{1}{4} (2x+1)^{-1} + frac{1}{8} (2x+1)^{-2}.$$






        share|cite|improve this answer









        $endgroup$



        1) Observe: $$x= frac 12 times left ( (2x + 1) -1right).$$



        2) Use this to obtain $$frac{x}{(2x+1)^3} = frac{1}{2} times frac{1}{(2x+1)^2} - frac 12 times frac{1}{(2x+1)^3}.$$



        3) Integrate to get



        $$ int frac{x}{(2x+1)^3} dx = -frac{1}{4} (2x+1)^{-1} + frac{1}{8} (2x+1)^{-2}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        FnacoolFnacool

        5,021511




        5,021511






























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            Tantalo (mitologio)

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            Ĉi tiu artikolo temas pri mitologio. Por aliaj signifoj vidu la paĝon Tantalo (apartigilo) . Tantalo , laŭ Gioacchino Assereto. Laŭ la helena mitologio, Tantalo (el la greka Τάνταλος , ŝancelanta aŭ tre malfeliĉa ) estis filo de Zeŭso kaj la oceanidino Pluto, laŭ la plej divastigata versio. Li estis reĝo de Frigio [1] aŭ de la monto Sipilo, en Lidio (Malgrandazio) [2] kaj post sia morto li suferis eternan torturadon ĉe la Tartaro, la plej profunda parto de Hadeso, rezervita por la plej perversaj homoj. Tamen, kelkaj rakontoj asertas ke li estis reĝo de Argo aŭ de Korinto [3] , aŭ ke li migris norden el la monto Sipilo por akiri la tronon de Paflagonio, el kie Ilo forpelis lin pro tio, ke Tantalo forkaptis kaj seksperfortis lian fraton Ganimedon [4] , kaj tiel ofendis la diojn. Verŝajne li ankaŭ regis urbon en okcidenta Anatolio, kiun oni nomis Tantalis [5] pro li, aŭ Sipilo , pro la monto sur kies flanko ankoraŭ staras ĝiaj ruinaĵoj [6] . Laŭ la geografiisto Strabo,...
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            Erzsébet Schaár

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            Erzsébet Schaár Memportreto de Erzsébet Schaár en Pécs Personaj informoj Naskiĝo 27-an de julio 1908 ( 1908-07-27 ) en Budafok Morto 29-an de aŭgusto 1975 ( 1975-08-29 ) (67-jara) en Budapeŝto Ŝtataneco Hungario Profesio skulptisto Familianoj Edz(in)o Tibor Vilt v   •   d   •   r Erzsébet Schaár [erĵEbet ŝAr], laŭ hungarlingve kutima sinsekvo Schaár Erzsébet estis hungara skulptisto. Ŝia edzo estis Tibor Vilt. Déryné Széppataki Róza en Miskolc fare de Erzsébet Schaár Erzsébet Schaár [1] naskiĝis la 27-an de julio 1908 en Budafok, ŝi mortis la 29-an de aŭgusto 1975 en Budapeŝto. Enhavo 1 Biografio 2 Elektitaj solaj ekspozicioj 3 Elektitaj kolektivaj ekspozicioj 4 Elektitaj skulptaĵoj 5 Elektitaj disĉiploj 6 Fontoj 7 Referencoj Biografio | Erzsébet Schaár studis en Industriarta Lernejo, poste en Reĝa Belarta Altlernejo (1924-1926). Ŝia ĉefinstruisto estis Zsig...
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