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
    41 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
    41 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
    41 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
    41 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
41 mins ago




$begingroup$
They have 1/18 as the answer @ThomasShelby
$endgroup$
– Jwan622
41 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
    18 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
      18 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
      18 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
      18 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
      18 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
    18 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
    18 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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            數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音...
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            When can things happen in Etherscan, such as the picture below?

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            2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  ...
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            格利澤436b

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            格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
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