Integral check. Is partial fractions the only way?












2












$begingroup$


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
    3 hours ago










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
















2












$begingroup$


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
    3 hours ago










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














2












2








2





$begingroup$


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 3 hours ago









Joshua Mundinger

2,5191028




2,5191028










asked 3 hours ago









Jwan622Jwan622

2,14111530




2,14111530












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










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


















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










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
















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




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












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




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










4 Answers
4






active

oldest

votes


















4












$begingroup$

Hint: Substitute $x=tfrac12 u -tfrac12$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You beat me to it.
    $endgroup$
    – randomgirl
    3 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
    51 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
    3 hours ago










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










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



















2












$begingroup$

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









$endgroup$





















    0












    $begingroup$

    $$I=int_{0}^{1}dfrac{x mathrm dx}{(2x+1)^3}implies 2I=int_{0}^{1}dfrac{2x+1-1}{(2x+1)^3}mathrm dx$$ $$2I=int_{0}^{1}biggl[dfrac{1}{(2x+1)^2}-dfrac{1}{(2x+1)^3}biggr] mathrm dx $$



    Make the substitution $begin{bmatrix}t \ mathrm dt end{bmatrix}=begin{bmatrix}2x+1 \ 2mathrm dxend{bmatrix}$ and use the general power rule for integrals:



    $$I=dfrac{1}{4}biggl[-dfrac{1}{t}+dfrac{1}{2t^2}biggr]_{1}^{3}=dfrac{1}{18}$$






    share|cite









    $endgroup$













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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Hint: Substitute $x=tfrac12 u -tfrac12$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        You beat me to it.
        $endgroup$
        – randomgirl
        3 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
        51 mins ago
















      4












      $begingroup$

      Hint: Substitute $x=tfrac12 u -tfrac12$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        You beat me to it.
        $endgroup$
        – randomgirl
        3 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
        51 mins ago














      4












      4








      4





      $begingroup$

      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 3 hours ago









      MPWMPW

      30.3k12157




      30.3k12157












      • $begingroup$
        You beat me to it.
        $endgroup$
        – randomgirl
        3 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
        51 mins ago


















      • $begingroup$
        You beat me to it.
        $endgroup$
        – randomgirl
        3 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
        51 mins ago
















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




      $begingroup$
      You beat me to it.
      $endgroup$
      – randomgirl
      3 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
      51 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
      51 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
        3 hours ago










      • $begingroup$
        @manooooh What do you mean? I don't understand.
        $endgroup$
        – Haris Gusic
        3 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
        3 hours ago










      • $begingroup$
        @manooooh What do you mean? I don't understand.
        $endgroup$
        – Haris Gusic
        3 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 3 hours ago









      Haris GusicHaris Gusic

      960116




      960116












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










      • $begingroup$
        @manooooh What do you mean? I don't understand.
        $endgroup$
        – Haris Gusic
        3 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
        3 hours ago










      • $begingroup$
        @manooooh What do you mean? I don't understand.
        $endgroup$
        – Haris Gusic
        3 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
      3 hours ago




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












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




      $begingroup$
      @manooooh What do you mean? I don't understand.
      $endgroup$
      – Haris Gusic
      3 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











      2












      $begingroup$

      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









      $endgroup$


















        2












        $begingroup$

        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









        $endgroup$
















          2












          2








          2





          $begingroup$

          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 3 hours ago









          FnacoolFnacool

          5,031511




          5,031511























              0












              $begingroup$

              $$I=int_{0}^{1}dfrac{x mathrm dx}{(2x+1)^3}implies 2I=int_{0}^{1}dfrac{2x+1-1}{(2x+1)^3}mathrm dx$$ $$2I=int_{0}^{1}biggl[dfrac{1}{(2x+1)^2}-dfrac{1}{(2x+1)^3}biggr] mathrm dx $$



              Make the substitution $begin{bmatrix}t \ mathrm dt end{bmatrix}=begin{bmatrix}2x+1 \ 2mathrm dxend{bmatrix}$ and use the general power rule for integrals:



              $$I=dfrac{1}{4}biggl[-dfrac{1}{t}+dfrac{1}{2t^2}biggr]_{1}^{3}=dfrac{1}{18}$$






              share|cite









              $endgroup$


















                0












                $begingroup$

                $$I=int_{0}^{1}dfrac{x mathrm dx}{(2x+1)^3}implies 2I=int_{0}^{1}dfrac{2x+1-1}{(2x+1)^3}mathrm dx$$ $$2I=int_{0}^{1}biggl[dfrac{1}{(2x+1)^2}-dfrac{1}{(2x+1)^3}biggr] mathrm dx $$



                Make the substitution $begin{bmatrix}t \ mathrm dt end{bmatrix}=begin{bmatrix}2x+1 \ 2mathrm dxend{bmatrix}$ and use the general power rule for integrals:



                $$I=dfrac{1}{4}biggl[-dfrac{1}{t}+dfrac{1}{2t^2}biggr]_{1}^{3}=dfrac{1}{18}$$






                share|cite









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $$I=int_{0}^{1}dfrac{x mathrm dx}{(2x+1)^3}implies 2I=int_{0}^{1}dfrac{2x+1-1}{(2x+1)^3}mathrm dx$$ $$2I=int_{0}^{1}biggl[dfrac{1}{(2x+1)^2}-dfrac{1}{(2x+1)^3}biggr] mathrm dx $$



                  Make the substitution $begin{bmatrix}t \ mathrm dt end{bmatrix}=begin{bmatrix}2x+1 \ 2mathrm dxend{bmatrix}$ and use the general power rule for integrals:



                  $$I=dfrac{1}{4}biggl[-dfrac{1}{t}+dfrac{1}{2t^2}biggr]_{1}^{3}=dfrac{1}{18}$$






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                  $endgroup$



                  $$I=int_{0}^{1}dfrac{x mathrm dx}{(2x+1)^3}implies 2I=int_{0}^{1}dfrac{2x+1-1}{(2x+1)^3}mathrm dx$$ $$2I=int_{0}^{1}biggl[dfrac{1}{(2x+1)^2}-dfrac{1}{(2x+1)^3}biggr] mathrm dx $$



                  Make the substitution $begin{bmatrix}t \ mathrm dt end{bmatrix}=begin{bmatrix}2x+1 \ 2mathrm dxend{bmatrix}$ and use the general power rule for integrals:



                  $$I=dfrac{1}{4}biggl[-dfrac{1}{t}+dfrac{1}{2t^2}biggr]_{1}^{3}=dfrac{1}{18}$$







                  share|cite












                  share|cite



                  share|cite










                  answered 4 mins ago









                  Paras KhoslaParas Khosla

                  733213




                  733213






























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