Integrating $intfrac1{x(x+2)}dx$ and $intfrac1{e^x+2}dx$. What am I doing wrong?












2












$begingroup$


I have the following 2 questions to integrate as part of my practice.



(i) $displaystyle intfrac1{x(x+2)}dx$



(ii) $displaystyle intfrac1{e^x+2}dx$



and I have tried it myself but I am not getting the answer required



(i) $displaystyle frac12ln x-frac12ln(x+2)+C$



(ii) $displaystyle frac12x-frac12ln(e^x+2)+C$



Below are my workings



(i) $displaystyle intfrac1{x(x+2)}dx;=;intfrac1{x^2+2x}dx;=;frac{ln(x^2+2x)}{2x+2} +C$



(ii) $displaystyle intfrac1{e^x+2}dx;=frac{ln(e^x+2)}{e^x} +C$



May I know what am I doing wrong such that my answer differs?






calculus integration indefinite-integrals







share|cite|improve this question











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  • 4




    $begingroup$
    Take the derivatives of the functions you got, what do you get? It should make you realize your error
    $endgroup$
    – Sorfosh
    Dec 2 '18 at 5:56










  • $begingroup$
    I didnt get back to the equation I want, but I don't understand why am I doing it wrong @Sorfosh
    $endgroup$
    – deviljones
    Dec 2 '18 at 6:00


















2












$begingroup$


I have the following 2 questions to integrate as part of my practice.



(i) $displaystyle intfrac1{x(x+2)}dx$



(ii) $displaystyle intfrac1{e^x+2}dx$



and I have tried it myself but I am not getting the answer required



(i) $displaystyle frac12ln x-frac12ln(x+2)+C$



(ii) $displaystyle frac12x-frac12ln(e^x+2)+C$



Below are my workings



(i) $displaystyle intfrac1{x(x+2)}dx;=;intfrac1{x^2+2x}dx;=;frac{ln(x^2+2x)}{2x+2} +C$



(ii) $displaystyle intfrac1{e^x+2}dx;=frac{ln(e^x+2)}{e^x} +C$



May I know what am I doing wrong such that my answer differs?






calculus integration indefinite-integrals







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Take the derivatives of the functions you got, what do you get? It should make you realize your error
    $endgroup$
    – Sorfosh
    Dec 2 '18 at 5:56










  • $begingroup$
    I didnt get back to the equation I want, but I don't understand why am I doing it wrong @Sorfosh
    $endgroup$
    – deviljones
    Dec 2 '18 at 6:00
















2












2








2





$begingroup$


I have the following 2 questions to integrate as part of my practice.



(i) $displaystyle intfrac1{x(x+2)}dx$



(ii) $displaystyle intfrac1{e^x+2}dx$



and I have tried it myself but I am not getting the answer required



(i) $displaystyle frac12ln x-frac12ln(x+2)+C$



(ii) $displaystyle frac12x-frac12ln(e^x+2)+C$



Below are my workings



(i) $displaystyle intfrac1{x(x+2)}dx;=;intfrac1{x^2+2x}dx;=;frac{ln(x^2+2x)}{2x+2} +C$



(ii) $displaystyle intfrac1{e^x+2}dx;=frac{ln(e^x+2)}{e^x} +C$



May I know what am I doing wrong such that my answer differs?






calculus integration indefinite-integrals







share|cite|improve this question











$endgroup$




I have the following 2 questions to integrate as part of my practice.



(i) $displaystyle intfrac1{x(x+2)}dx$



(ii) $displaystyle intfrac1{e^x+2}dx$



and I have tried it myself but I am not getting the answer required



(i) $displaystyle frac12ln x-frac12ln(x+2)+C$



(ii) $displaystyle frac12x-frac12ln(e^x+2)+C$



Below are my workings



(i) $displaystyle intfrac1{x(x+2)}dx;=;intfrac1{x^2+2x}dx;=;frac{ln(x^2+2x)}{2x+2} +C$



(ii) $displaystyle intfrac1{e^x+2}dx;=frac{ln(e^x+2)}{e^x} +C$



May I know what am I doing wrong such that my answer differs?






calculus integration indefinite-integrals




calculus integration indefinite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 5:53









Blue

47.8k870152




47.8k870152










asked Dec 2 '18 at 5:49









deviljonesdeviljones

636




636








  • 4




    $begingroup$
    Take the derivatives of the functions you got, what do you get? It should make you realize your error
    $endgroup$
    – Sorfosh
    Dec 2 '18 at 5:56










  • $begingroup$
    I didnt get back to the equation I want, but I don't understand why am I doing it wrong @Sorfosh
    $endgroup$
    – deviljones
    Dec 2 '18 at 6:00
















  • 4




    $begingroup$
    Take the derivatives of the functions you got, what do you get? It should make you realize your error
    $endgroup$
    – Sorfosh
    Dec 2 '18 at 5:56










  • $begingroup$
    I didnt get back to the equation I want, but I don't understand why am I doing it wrong @Sorfosh
    $endgroup$
    – deviljones
    Dec 2 '18 at 6:00










4




4




$begingroup$
Take the derivatives of the functions you got, what do you get? It should make you realize your error
$endgroup$
– Sorfosh
Dec 2 '18 at 5:56




$begingroup$
Take the derivatives of the functions you got, what do you get? It should make you realize your error
$endgroup$
– Sorfosh
Dec 2 '18 at 5:56












$begingroup$
I didnt get back to the equation I want, but I don't understand why am I doing it wrong @Sorfosh
$endgroup$
– deviljones
Dec 2 '18 at 6:00






$begingroup$
I didnt get back to the equation I want, but I don't understand why am I doing it wrong @Sorfosh
$endgroup$
– deviljones
Dec 2 '18 at 6:00












3 Answers
3






active

oldest

votes


















1












$begingroup$

For i):



With the constant rule 'trick', what you are effectively doing is:
$$int frac{1}{x^2+2x} dx$$



When $u = x^2+2x, du = 2x+2 dx, dx = frac{du}{2x+2}$, we have:



$$int frac{1}{u(2x+2)} du, $$



and since you cannot get the $2x+2$ part to be in terms of $u$, this 'trick' does not work.



What you can do is to use partial fractions:
$$frac{A}{x} + frac{B}{x+2} = frac{1}{x(x+2)}$$
$$A(x+2) + Bx = 1 tag{1}label{eq1}$$
$$(A+B)x + 2A = 1 tag{2}label{eq2}$$



This is an identity – it will hold for any $x$. So when $x = 0, A = frac{1}{2}$ from equation $2$, and when $x = -2$, $B = -frac{1}{2}$ from equation $1$.



Can you integrate it now?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    yup,, thanks man
    $endgroup$
    – deviljones
    Dec 2 '18 at 6:28










  • $begingroup$
    @deviljones No problem!
    $endgroup$
    – Toby Mak
    Dec 2 '18 at 7:07



















3












$begingroup$

For i) use partial fractions; note that $$frac{1}{x(x+2)}=frac{frac 12}{x} -frac{frac12}{x+2}$$ So your integral becomes:
$$int{frac{1}{x(x+2)}dx}=frac12bigg[int frac 1x dx-int frac{1}{x+2}dxbigg]$$



For the second, substitute $$u=e^xto dx = frac{du}{u}$$



Then your integral becomes:



$$intfrac{1}{e^x+2}dx=int{frac{1}{u(u+2)}du}$$ which we just solved in i)






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    You're missing a $dx$ in the last line.
    $endgroup$
    – Toby Mak
    Dec 2 '18 at 7:06










  • $begingroup$
    Whoops. Thanks.
    $endgroup$
    – Rhys Hughes
    Dec 2 '18 at 18:53



















0












$begingroup$

The substitution rule is $$int f(g(x)) , g'(x) ,{rm d}x = int f(u) ,{rm d}u.$$ In particular, when $f(x) = 1/x$, $$int frac{g'(x)}{g(x)},{rm d}x = int frac{{rm d}u}{u} = log u + C = log(g(x)) + C.$$



What you are trying to do is to "dividing both sides by the derivative of $g$": $$int frac{1}{g(x)},{rm d}x = frac{log(g(x))}{g'(x)} + C. tag{OP's wrong method}$$ As the first comment suggests, differentiating both sides with respect to $x$ allows you to see the error.






share|cite|improve this answer









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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    For i):



    With the constant rule 'trick', what you are effectively doing is:
    $$int frac{1}{x^2+2x} dx$$



    When $u = x^2+2x, du = 2x+2 dx, dx = frac{du}{2x+2}$, we have:



    $$int frac{1}{u(2x+2)} du, $$



    and since you cannot get the $2x+2$ part to be in terms of $u$, this 'trick' does not work.



    What you can do is to use partial fractions:
    $$frac{A}{x} + frac{B}{x+2} = frac{1}{x(x+2)}$$
    $$A(x+2) + Bx = 1 tag{1}label{eq1}$$
    $$(A+B)x + 2A = 1 tag{2}label{eq2}$$



    This is an identity – it will hold for any $x$. So when $x = 0, A = frac{1}{2}$ from equation $2$, and when $x = -2$, $B = -frac{1}{2}$ from equation $1$.



    Can you integrate it now?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      yup,, thanks man
      $endgroup$
      – deviljones
      Dec 2 '18 at 6:28










    • $begingroup$
      @deviljones No problem!
      $endgroup$
      – Toby Mak
      Dec 2 '18 at 7:07
















    1












    $begingroup$

    For i):



    With the constant rule 'trick', what you are effectively doing is:
    $$int frac{1}{x^2+2x} dx$$



    When $u = x^2+2x, du = 2x+2 dx, dx = frac{du}{2x+2}$, we have:



    $$int frac{1}{u(2x+2)} du, $$



    and since you cannot get the $2x+2$ part to be in terms of $u$, this 'trick' does not work.



    What you can do is to use partial fractions:
    $$frac{A}{x} + frac{B}{x+2} = frac{1}{x(x+2)}$$
    $$A(x+2) + Bx = 1 tag{1}label{eq1}$$
    $$(A+B)x + 2A = 1 tag{2}label{eq2}$$



    This is an identity – it will hold for any $x$. So when $x = 0, A = frac{1}{2}$ from equation $2$, and when $x = -2$, $B = -frac{1}{2}$ from equation $1$.



    Can you integrate it now?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      yup,, thanks man
      $endgroup$
      – deviljones
      Dec 2 '18 at 6:28










    • $begingroup$
      @deviljones No problem!
      $endgroup$
      – Toby Mak
      Dec 2 '18 at 7:07














    1












    1








    1





    $begingroup$

    For i):



    With the constant rule 'trick', what you are effectively doing is:
    $$int frac{1}{x^2+2x} dx$$



    When $u = x^2+2x, du = 2x+2 dx, dx = frac{du}{2x+2}$, we have:



    $$int frac{1}{u(2x+2)} du, $$



    and since you cannot get the $2x+2$ part to be in terms of $u$, this 'trick' does not work.



    What you can do is to use partial fractions:
    $$frac{A}{x} + frac{B}{x+2} = frac{1}{x(x+2)}$$
    $$A(x+2) + Bx = 1 tag{1}label{eq1}$$
    $$(A+B)x + 2A = 1 tag{2}label{eq2}$$



    This is an identity – it will hold for any $x$. So when $x = 0, A = frac{1}{2}$ from equation $2$, and when $x = -2$, $B = -frac{1}{2}$ from equation $1$.



    Can you integrate it now?






    share|cite|improve this answer









    $endgroup$



    For i):



    With the constant rule 'trick', what you are effectively doing is:
    $$int frac{1}{x^2+2x} dx$$



    When $u = x^2+2x, du = 2x+2 dx, dx = frac{du}{2x+2}$, we have:



    $$int frac{1}{u(2x+2)} du, $$



    and since you cannot get the $2x+2$ part to be in terms of $u$, this 'trick' does not work.



    What you can do is to use partial fractions:
    $$frac{A}{x} + frac{B}{x+2} = frac{1}{x(x+2)}$$
    $$A(x+2) + Bx = 1 tag{1}label{eq1}$$
    $$(A+B)x + 2A = 1 tag{2}label{eq2}$$



    This is an identity – it will hold for any $x$. So when $x = 0, A = frac{1}{2}$ from equation $2$, and when $x = -2$, $B = -frac{1}{2}$ from equation $1$.



    Can you integrate it now?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 2 '18 at 6:09









    Toby MakToby Mak

    3,37811128




    3,37811128












    • $begingroup$
      yup,, thanks man
      $endgroup$
      – deviljones
      Dec 2 '18 at 6:28










    • $begingroup$
      @deviljones No problem!
      $endgroup$
      – Toby Mak
      Dec 2 '18 at 7:07


















    • $begingroup$
      yup,, thanks man
      $endgroup$
      – deviljones
      Dec 2 '18 at 6:28










    • $begingroup$
      @deviljones No problem!
      $endgroup$
      – Toby Mak
      Dec 2 '18 at 7:07
















    $begingroup$
    yup,, thanks man
    $endgroup$
    – deviljones
    Dec 2 '18 at 6:28




    $begingroup$
    yup,, thanks man
    $endgroup$
    – deviljones
    Dec 2 '18 at 6:28












    $begingroup$
    @deviljones No problem!
    $endgroup$
    – Toby Mak
    Dec 2 '18 at 7:07




    $begingroup$
    @deviljones No problem!
    $endgroup$
    – Toby Mak
    Dec 2 '18 at 7:07











    3












    $begingroup$

    For i) use partial fractions; note that $$frac{1}{x(x+2)}=frac{frac 12}{x} -frac{frac12}{x+2}$$ So your integral becomes:
    $$int{frac{1}{x(x+2)}dx}=frac12bigg[int frac 1x dx-int frac{1}{x+2}dxbigg]$$



    For the second, substitute $$u=e^xto dx = frac{du}{u}$$



    Then your integral becomes:



    $$intfrac{1}{e^x+2}dx=int{frac{1}{u(u+2)}du}$$ which we just solved in i)






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      You're missing a $dx$ in the last line.
      $endgroup$
      – Toby Mak
      Dec 2 '18 at 7:06










    • $begingroup$
      Whoops. Thanks.
      $endgroup$
      – Rhys Hughes
      Dec 2 '18 at 18:53
















    3












    $begingroup$

    For i) use partial fractions; note that $$frac{1}{x(x+2)}=frac{frac 12}{x} -frac{frac12}{x+2}$$ So your integral becomes:
    $$int{frac{1}{x(x+2)}dx}=frac12bigg[int frac 1x dx-int frac{1}{x+2}dxbigg]$$



    For the second, substitute $$u=e^xto dx = frac{du}{u}$$



    Then your integral becomes:



    $$intfrac{1}{e^x+2}dx=int{frac{1}{u(u+2)}du}$$ which we just solved in i)






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      You're missing a $dx$ in the last line.
      $endgroup$
      – Toby Mak
      Dec 2 '18 at 7:06










    • $begingroup$
      Whoops. Thanks.
      $endgroup$
      – Rhys Hughes
      Dec 2 '18 at 18:53














    3












    3








    3





    $begingroup$

    For i) use partial fractions; note that $$frac{1}{x(x+2)}=frac{frac 12}{x} -frac{frac12}{x+2}$$ So your integral becomes:
    $$int{frac{1}{x(x+2)}dx}=frac12bigg[int frac 1x dx-int frac{1}{x+2}dxbigg]$$



    For the second, substitute $$u=e^xto dx = frac{du}{u}$$



    Then your integral becomes:



    $$intfrac{1}{e^x+2}dx=int{frac{1}{u(u+2)}du}$$ which we just solved in i)






    share|cite|improve this answer











    $endgroup$



    For i) use partial fractions; note that $$frac{1}{x(x+2)}=frac{frac 12}{x} -frac{frac12}{x+2}$$ So your integral becomes:
    $$int{frac{1}{x(x+2)}dx}=frac12bigg[int frac 1x dx-int frac{1}{x+2}dxbigg]$$



    For the second, substitute $$u=e^xto dx = frac{du}{u}$$



    Then your integral becomes:



    $$intfrac{1}{e^x+2}dx=int{frac{1}{u(u+2)}du}$$ which we just solved in i)







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 2 '18 at 18:52

























    answered Dec 2 '18 at 6:31









    Rhys HughesRhys Hughes

    5,1961427




    5,1961427








    • 1




      $begingroup$
      You're missing a $dx$ in the last line.
      $endgroup$
      – Toby Mak
      Dec 2 '18 at 7:06










    • $begingroup$
      Whoops. Thanks.
      $endgroup$
      – Rhys Hughes
      Dec 2 '18 at 18:53














    • 1




      $begingroup$
      You're missing a $dx$ in the last line.
      $endgroup$
      – Toby Mak
      Dec 2 '18 at 7:06










    • $begingroup$
      Whoops. Thanks.
      $endgroup$
      – Rhys Hughes
      Dec 2 '18 at 18:53








    1




    1




    $begingroup$
    You're missing a $dx$ in the last line.
    $endgroup$
    – Toby Mak
    Dec 2 '18 at 7:06




    $begingroup$
    You're missing a $dx$ in the last line.
    $endgroup$
    – Toby Mak
    Dec 2 '18 at 7:06












    $begingroup$
    Whoops. Thanks.
    $endgroup$
    – Rhys Hughes
    Dec 2 '18 at 18:53




    $begingroup$
    Whoops. Thanks.
    $endgroup$
    – Rhys Hughes
    Dec 2 '18 at 18:53











    0












    $begingroup$

    The substitution rule is $$int f(g(x)) , g'(x) ,{rm d}x = int f(u) ,{rm d}u.$$ In particular, when $f(x) = 1/x$, $$int frac{g'(x)}{g(x)},{rm d}x = int frac{{rm d}u}{u} = log u + C = log(g(x)) + C.$$



    What you are trying to do is to "dividing both sides by the derivative of $g$": $$int frac{1}{g(x)},{rm d}x = frac{log(g(x))}{g'(x)} + C. tag{OP's wrong method}$$ As the first comment suggests, differentiating both sides with respect to $x$ allows you to see the error.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The substitution rule is $$int f(g(x)) , g'(x) ,{rm d}x = int f(u) ,{rm d}u.$$ In particular, when $f(x) = 1/x$, $$int frac{g'(x)}{g(x)},{rm d}x = int frac{{rm d}u}{u} = log u + C = log(g(x)) + C.$$



      What you are trying to do is to "dividing both sides by the derivative of $g$": $$int frac{1}{g(x)},{rm d}x = frac{log(g(x))}{g'(x)} + C. tag{OP's wrong method}$$ As the first comment suggests, differentiating both sides with respect to $x$ allows you to see the error.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The substitution rule is $$int f(g(x)) , g'(x) ,{rm d}x = int f(u) ,{rm d}u.$$ In particular, when $f(x) = 1/x$, $$int frac{g'(x)}{g(x)},{rm d}x = int frac{{rm d}u}{u} = log u + C = log(g(x)) + C.$$



        What you are trying to do is to "dividing both sides by the derivative of $g$": $$int frac{1}{g(x)},{rm d}x = frac{log(g(x))}{g'(x)} + C. tag{OP's wrong method}$$ As the first comment suggests, differentiating both sides with respect to $x$ allows you to see the error.






        share|cite|improve this answer









        $endgroup$



        The substitution rule is $$int f(g(x)) , g'(x) ,{rm d}x = int f(u) ,{rm d}u.$$ In particular, when $f(x) = 1/x$, $$int frac{g'(x)}{g(x)},{rm d}x = int frac{{rm d}u}{u} = log u + C = log(g(x)) + C.$$



        What you are trying to do is to "dividing both sides by the derivative of $g$": $$int frac{1}{g(x)},{rm d}x = frac{log(g(x))}{g'(x)} + C. tag{OP's wrong method}$$ As the first comment suggests, differentiating both sides with respect to $x$ allows you to see the error.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 6:09









        GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會

        12.8k72445




        12.8k72445






























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