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
    42 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
    42 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
    42 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
    42 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
42 mins ago




$begingroup$
They have 1/18 as the answer @ThomasShelby
$endgroup$
– Jwan622
42 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
    19 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
      19 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
      19 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
      19 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
      19 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
    19 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
    19 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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            How to solve the problem of ntp “Unable to contact time server” from KDE?

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            1 0 I want to set date and time automatically but this error appears when I try to .. I'm using openSUSE, I tried different servers but it didn't work either, help please! It doesn't seem to be a network problem: $ ping pool.ntp.org 64 bytes from 41.78.128.17: icmp_seq=1 ttl=41 time=260 ms To the question of whether ntpd is installed and running: $ ps -C ntpd PID TTY TIME CMD i.e. there's no ntpd process running. $ rpm -qa | grep ntp yast2-ntp-client-3.1.12-1.7.noarch ntp-4.2.6p5-25.2.1.i586 Further information from the comments: $ sudo ntpdate pool.ntp.org 25 Sep 19:41:51 ntpdate[3162]: no server suitable for synchronization found networking kde opensuse ntp ntpd sh
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            ASUS Zenbook UX433/UX333 — Configure Touchpad-embedded numpad on Linux

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            0 1 I have an ASUS Zenbook UX433,on which ASUS decided to innovate by embedding a Numpad inside the touchpad (see picture 1). Toggling between classic touchpad mode and numpad (also turns the touchpad LEDs on) can be done through the golden button in the top-right corner of the touchpad. However, on my Manjaro arch Linux config, this behavior does not work by default and I can only use the touchpad, no numpad. Has anyone been able to configure the numpad to work on Linux, or have any idea how this could be achieved? EDIT1: I have had a look at the official Windows drivers for the UX433FA TouchPad, but was not able to make anything of it. I tried installing them on Linux using tools like NDISWrapper, but they do not detect the TouchPad/Numpad hardware, and my understanding of drivers is too limited.
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