Evaluate the intgral $intfracdxx^2(1-x^2)$ (solution verification)

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3
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I have to solve the following integral
$$intfracdxx^2(1-x^2)$$



What I've got:
beginsplit
intfracdxx^2(1-x^2) &=intfrac(1-x^2+x^2)dxx^2(1-x^2)\
&=intfracdxx^2+intfracdx1-x^2\
&=intfracdxx^2+intfracdx(1+(xi)^2)\
&=-x^-1+arctanxi+C
endsplit



Is this correct?



Thanks in advance!




calculus integration proof-verification




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  • 1




    I would suggest partial fraction decomposition for the last integral instead of setting $x=xi$. You cannot just change a variable without changing the differential too but it should work aswell.
    – mrtaurho
    Aug 2 at 11:35







  • 1




    As user 108128 already noted, you missed a factor of $-i$ on the arctangent. In fact we can write $-ixarctan ix$ as a hyperbolic arctangent, $operatornameartanh x$.
    – J.G.
    Aug 2 at 11:57











  • I think it's good that you came up with your own method, but you probably went the wrong way on the $frac11 - x^2$ part, which can be done different ways google.com/…
    – Cato
    Aug 2 at 12:58















up vote
3
down vote

favorite












I have to solve the following integral
$$intfracdxx^2(1-x^2)$$



What I've got:
beginsplit
intfracdxx^2(1-x^2) &=intfrac(1-x^2+x^2)dxx^2(1-x^2)\
&=intfracdxx^2+intfracdx1-x^2\
&=intfracdxx^2+intfracdx(1+(xi)^2)\
&=-x^-1+arctanxi+C
endsplit



Is this correct?



Thanks in advance!




calculus integration proof-verification




share|cite|improve this question

















  • 1




    I would suggest partial fraction decomposition for the last integral instead of setting $x=xi$. You cannot just change a variable without changing the differential too but it should work aswell.
    – mrtaurho
    Aug 2 at 11:35







  • 1




    As user 108128 already noted, you missed a factor of $-i$ on the arctangent. In fact we can write $-ixarctan ix$ as a hyperbolic arctangent, $operatornameartanh x$.
    – J.G.
    Aug 2 at 11:57











  • I think it's good that you came up with your own method, but you probably went the wrong way on the $frac11 - x^2$ part, which can be done different ways google.com/…
    – Cato
    Aug 2 at 12:58













up vote
3
down vote

favorite









up vote
3
down vote

favorite











I have to solve the following integral
$$intfracdxx^2(1-x^2)$$



What I've got:
beginsplit
intfracdxx^2(1-x^2) &=intfrac(1-x^2+x^2)dxx^2(1-x^2)\
&=intfracdxx^2+intfracdx1-x^2\
&=intfracdxx^2+intfracdx(1+(xi)^2)\
&=-x^-1+arctanxi+C
endsplit



Is this correct?



Thanks in advance!




calculus integration proof-verification




share|cite|improve this question













I have to solve the following integral
$$intfracdxx^2(1-x^2)$$



What I've got:
beginsplit
intfracdxx^2(1-x^2) &=intfrac(1-x^2+x^2)dxx^2(1-x^2)\
&=intfracdxx^2+intfracdx1-x^2\
&=intfracdxx^2+intfracdx(1+(xi)^2)\
&=-x^-1+arctanxi+C
endsplit



Is this correct?



Thanks in advance!





calculus integration proof-verification





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 2 at 14:57









John Ma

37.5k93669




37.5k93669









asked Aug 2 at 11:31









Nikola

690617




690617







  • 1




    I would suggest partial fraction decomposition for the last integral instead of setting $x=xi$. You cannot just change a variable without changing the differential too but it should work aswell.
    – mrtaurho
    Aug 2 at 11:35







  • 1




    As user 108128 already noted, you missed a factor of $-i$ on the arctangent. In fact we can write $-ixarctan ix$ as a hyperbolic arctangent, $operatornameartanh x$.
    – J.G.
    Aug 2 at 11:57











  • I think it's good that you came up with your own method, but you probably went the wrong way on the $frac11 - x^2$ part, which can be done different ways google.com/…
    – Cato
    Aug 2 at 12:58













  • 1




    I would suggest partial fraction decomposition for the last integral instead of setting $x=xi$. You cannot just change a variable without changing the differential too but it should work aswell.
    – mrtaurho
    Aug 2 at 11:35







  • 1




    As user 108128 already noted, you missed a factor of $-i$ on the arctangent. In fact we can write $-ixarctan ix$ as a hyperbolic arctangent, $operatornameartanh x$.
    – J.G.
    Aug 2 at 11:57











  • I think it's good that you came up with your own method, but you probably went the wrong way on the $frac11 - x^2$ part, which can be done different ways google.com/…
    – Cato
    Aug 2 at 12:58








1




1




I would suggest partial fraction decomposition for the last integral instead of setting $x=xi$. You cannot just change a variable without changing the differential too but it should work aswell.
– mrtaurho
Aug 2 at 11:35





I would suggest partial fraction decomposition for the last integral instead of setting $x=xi$. You cannot just change a variable without changing the differential too but it should work aswell.
– mrtaurho
Aug 2 at 11:35





1




1




As user 108128 already noted, you missed a factor of $-i$ on the arctangent. In fact we can write $-ixarctan ix$ as a hyperbolic arctangent, $operatornameartanh x$.
– J.G.
Aug 2 at 11:57





As user 108128 already noted, you missed a factor of $-i$ on the arctangent. In fact we can write $-ixarctan ix$ as a hyperbolic arctangent, $operatornameartanh x$.
– J.G.
Aug 2 at 11:57













I think it's good that you came up with your own method, but you probably went the wrong way on the $frac11 - x^2$ part, which can be done different ways google.com/…
– Cato
Aug 2 at 12:58





I think it's good that you came up with your own method, but you probably went the wrong way on the $frac11 - x^2$ part, which can be done different ways google.com/…
– Cato
Aug 2 at 12:58











5 Answers
5






active

oldest

votes

















up vote
3
down vote



accepted










Don't forget $dfrac1i$ before $arctan$
$$intfracdxx^2(1-x^2)=-x^-1+dfrac1iarctanxi+C$$
Also better to write
$$intdfrac11-x^2dx=dfrac12lndfrac1+x1-x+C$$






share|cite|improve this answer





















  • What problems there might be using the imaginary part? Is it a matter of preference?
    – Nikola
    Aug 2 at 11:51










  • Well, you say "its better to write". Why is that?
    – Nikola
    Aug 2 at 11:57






  • 1




    In real analysis we should avoid using imaginary part as possible, and it's better use functions with one variable $x$, and leave $i$ for complex analyysis functions.
    – user 108128
    Aug 2 at 12:02

















up vote
3
down vote













Another method using the self-similar substitution



$$x=frac1-t1+t$$



transforms the integral to
$$-intfrac2left( t-1 right)^2+frac12tdt$$






share|cite|improve this answer






























    up vote
    2
    down vote













    Are you sure that a non-real answer is acceptable? Even if it is, what you should have got as a primitive of $frac11-x^2$ should have been $-iarctan(xi)$. But I suggest that you do$$frac11-x^2=frac1/21-x+frac1/21+x$$instead.






    share|cite|improve this answer




























      up vote
      2
      down vote













      $x = tanh(u)$ then $dx = operatornamesech^2(u) du$
      $$
      beginalign
      int fracdxx^2(1-x^2)
      &= int fracoperatornamesech^2(u) dutanh^2(u)(1-tanh^2(u))\
      &= int fracoperatornamesech^2(u) dutanh^2(u)(operatornamesech^2(u))\
      &= int fracdutanh^2(u)\
      &= int fraccosh^2(u)sinh^2(u) du\
      &= int (1 + operatornamecsch^2(u))du\
      &= u - coth(u)\
      &= operatornamearctanh(x) - frac1x + C
      endalign
      $$






      share|cite|improve this answer























      • I've edited yours, if you don't want, so do roll back.
        – user 108128
        Aug 2 at 12:44










      • I think the original questioners partial fraction method was better - apart from missing the factorization of $1 - x^2$ - it was probably easier. I once integrated $frac11 - x^2$ the same sort of way and handed it in to someone
        – Cato
        Aug 2 at 13:00


















      up vote
      1
      down vote













      Note that, if $f(x)=arctan(xi)$, then
      $$
      f'(x)=ifrac11+(xi)^2=fraci1-x^2
      $$
      that's not the required derivative; you are missing $1/i=-i$, so you could write
      $$
      -frac1x-iarctan(xi)
      $$



      I'd prefer to avoid complex functions. With partial fractions, we set
      $$
      frac1x^2(1-x)^2=
      fracax+fracbx^2+fraccx-1+fracdx+1
      $$
      Removing the denominators leads to
      $$
      ax(x^2-1)+b(x^2-1)+cx^2(x+1)+dx^2(x-1)=-1
      $$
      With $x=0$, we get $-b=-1$; with $x=1$, we get $2c=-1$; with $x=-1$ we get $-2d=-1$. It remains to find $a$, but this is easily seen to be $0$.



      Therefore
      $$
      intfrac1x^2(1-x^2),dx=
      intleft(
      frac1x^2-frac12frac1x-1+frac12frac1x+1
      right),dx=-frac1x+frac12logleft|fracx+1x-1right|+c
      $$






      share|cite|improve this answer























      • +1. Good illustration. end of solution is $frac12$ behind $log$.
        – user 108128
        Aug 2 at 13:38











      • @user108128 Thanks for noting
        – egreg
        Aug 2 at 14:14










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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      3
      down vote



      accepted










      Don't forget $dfrac1i$ before $arctan$
      $$intfracdxx^2(1-x^2)=-x^-1+dfrac1iarctanxi+C$$
      Also better to write
      $$intdfrac11-x^2dx=dfrac12lndfrac1+x1-x+C$$






      share|cite|improve this answer





















      • What problems there might be using the imaginary part? Is it a matter of preference?
        – Nikola
        Aug 2 at 11:51










      • Well, you say "its better to write". Why is that?
        – Nikola
        Aug 2 at 11:57






      • 1




        In real analysis we should avoid using imaginary part as possible, and it's better use functions with one variable $x$, and leave $i$ for complex analyysis functions.
        – user 108128
        Aug 2 at 12:02














      up vote
      3
      down vote



      accepted










      Don't forget $dfrac1i$ before $arctan$
      $$intfracdxx^2(1-x^2)=-x^-1+dfrac1iarctanxi+C$$
      Also better to write
      $$intdfrac11-x^2dx=dfrac12lndfrac1+x1-x+C$$






      share|cite|improve this answer





















      • What problems there might be using the imaginary part? Is it a matter of preference?
        – Nikola
        Aug 2 at 11:51










      • Well, you say "its better to write". Why is that?
        – Nikola
        Aug 2 at 11:57






      • 1




        In real analysis we should avoid using imaginary part as possible, and it's better use functions with one variable $x$, and leave $i$ for complex analyysis functions.
        – user 108128
        Aug 2 at 12:02












      up vote
      3
      down vote



      accepted







      up vote
      3
      down vote



      accepted






      Don't forget $dfrac1i$ before $arctan$
      $$intfracdxx^2(1-x^2)=-x^-1+dfrac1iarctanxi+C$$
      Also better to write
      $$intdfrac11-x^2dx=dfrac12lndfrac1+x1-x+C$$






      share|cite|improve this answer













      Don't forget $dfrac1i$ before $arctan$
      $$intfracdxx^2(1-x^2)=-x^-1+dfrac1iarctanxi+C$$
      Also better to write
      $$intdfrac11-x^2dx=dfrac12lndfrac1+x1-x+C$$







      share|cite|improve this answer













      share|cite|improve this answer



      share|cite|improve this answer











      answered Aug 2 at 11:35









      user 108128

      18.8k41544




      18.8k41544











      • What problems there might be using the imaginary part? Is it a matter of preference?
        – Nikola
        Aug 2 at 11:51










      • Well, you say "its better to write". Why is that?
        – Nikola
        Aug 2 at 11:57






      • 1




        In real analysis we should avoid using imaginary part as possible, and it's better use functions with one variable $x$, and leave $i$ for complex analyysis functions.
        – user 108128
        Aug 2 at 12:02
















      • What problems there might be using the imaginary part? Is it a matter of preference?
        – Nikola
        Aug 2 at 11:51










      • Well, you say "its better to write". Why is that?
        – Nikola
        Aug 2 at 11:57






      • 1




        In real analysis we should avoid using imaginary part as possible, and it's better use functions with one variable $x$, and leave $i$ for complex analyysis functions.
        – user 108128
        Aug 2 at 12:02















      What problems there might be using the imaginary part? Is it a matter of preference?
      – Nikola
      Aug 2 at 11:51




      What problems there might be using the imaginary part? Is it a matter of preference?
      – Nikola
      Aug 2 at 11:51












      Well, you say "its better to write". Why is that?
      – Nikola
      Aug 2 at 11:57




      Well, you say "its better to write". Why is that?
      – Nikola
      Aug 2 at 11:57




      1




      1




      In real analysis we should avoid using imaginary part as possible, and it's better use functions with one variable $x$, and leave $i$ for complex analyysis functions.
      – user 108128
      Aug 2 at 12:02




      In real analysis we should avoid using imaginary part as possible, and it's better use functions with one variable $x$, and leave $i$ for complex analyysis functions.
      – user 108128
      Aug 2 at 12:02










      up vote
      3
      down vote













      Another method using the self-similar substitution



      $$x=frac1-t1+t$$



      transforms the integral to
      $$-intfrac2left( t-1 right)^2+frac12tdt$$






      share|cite|improve this answer



























        up vote
        3
        down vote













        Another method using the self-similar substitution



        $$x=frac1-t1+t$$



        transforms the integral to
        $$-intfrac2left( t-1 right)^2+frac12tdt$$






        share|cite|improve this answer

























          up vote
          3
          down vote










          up vote
          3
          down vote









          Another method using the self-similar substitution



          $$x=frac1-t1+t$$



          transforms the integral to
          $$-intfrac2left( t-1 right)^2+frac12tdt$$






          share|cite|improve this answer















          Another method using the self-similar substitution



          $$x=frac1-t1+t$$



          transforms the integral to
          $$-intfrac2left( t-1 right)^2+frac12tdt$$







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 2 at 22:38


























          answered Aug 2 at 22:23









          Vincent Law

          1,192110




          1,192110




















              up vote
              2
              down vote













              Are you sure that a non-real answer is acceptable? Even if it is, what you should have got as a primitive of $frac11-x^2$ should have been $-iarctan(xi)$. But I suggest that you do$$frac11-x^2=frac1/21-x+frac1/21+x$$instead.






              share|cite|improve this answer

























                up vote
                2
                down vote













                Are you sure that a non-real answer is acceptable? Even if it is, what you should have got as a primitive of $frac11-x^2$ should have been $-iarctan(xi)$. But I suggest that you do$$frac11-x^2=frac1/21-x+frac1/21+x$$instead.






                share|cite|improve this answer























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  Are you sure that a non-real answer is acceptable? Even if it is, what you should have got as a primitive of $frac11-x^2$ should have been $-iarctan(xi)$. But I suggest that you do$$frac11-x^2=frac1/21-x+frac1/21+x$$instead.






                  share|cite|improve this answer













                  Are you sure that a non-real answer is acceptable? Even if it is, what you should have got as a primitive of $frac11-x^2$ should have been $-iarctan(xi)$. But I suggest that you do$$frac11-x^2=frac1/21-x+frac1/21+x$$instead.







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Aug 2 at 11:35









                  José Carlos Santos

                  112k1696172




                  112k1696172




















                      up vote
                      2
                      down vote













                      $x = tanh(u)$ then $dx = operatornamesech^2(u) du$
                      $$
                      beginalign
                      int fracdxx^2(1-x^2)
                      &= int fracoperatornamesech^2(u) dutanh^2(u)(1-tanh^2(u))\
                      &= int fracoperatornamesech^2(u) dutanh^2(u)(operatornamesech^2(u))\
                      &= int fracdutanh^2(u)\
                      &= int fraccosh^2(u)sinh^2(u) du\
                      &= int (1 + operatornamecsch^2(u))du\
                      &= u - coth(u)\
                      &= operatornamearctanh(x) - frac1x + C
                      endalign
                      $$






                      share|cite|improve this answer























                      • I've edited yours, if you don't want, so do roll back.
                        – user 108128
                        Aug 2 at 12:44










                      • I think the original questioners partial fraction method was better - apart from missing the factorization of $1 - x^2$ - it was probably easier. I once integrated $frac11 - x^2$ the same sort of way and handed it in to someone
                        – Cato
                        Aug 2 at 13:00















                      up vote
                      2
                      down vote













                      $x = tanh(u)$ then $dx = operatornamesech^2(u) du$
                      $$
                      beginalign
                      int fracdxx^2(1-x^2)
                      &= int fracoperatornamesech^2(u) dutanh^2(u)(1-tanh^2(u))\
                      &= int fracoperatornamesech^2(u) dutanh^2(u)(operatornamesech^2(u))\
                      &= int fracdutanh^2(u)\
                      &= int fraccosh^2(u)sinh^2(u) du\
                      &= int (1 + operatornamecsch^2(u))du\
                      &= u - coth(u)\
                      &= operatornamearctanh(x) - frac1x + C
                      endalign
                      $$






                      share|cite|improve this answer























                      • I've edited yours, if you don't want, so do roll back.
                        – user 108128
                        Aug 2 at 12:44










                      • I think the original questioners partial fraction method was better - apart from missing the factorization of $1 - x^2$ - it was probably easier. I once integrated $frac11 - x^2$ the same sort of way and handed it in to someone
                        – Cato
                        Aug 2 at 13:00













                      up vote
                      2
                      down vote










                      up vote
                      2
                      down vote









                      $x = tanh(u)$ then $dx = operatornamesech^2(u) du$
                      $$
                      beginalign
                      int fracdxx^2(1-x^2)
                      &= int fracoperatornamesech^2(u) dutanh^2(u)(1-tanh^2(u))\
                      &= int fracoperatornamesech^2(u) dutanh^2(u)(operatornamesech^2(u))\
                      &= int fracdutanh^2(u)\
                      &= int fraccosh^2(u)sinh^2(u) du\
                      &= int (1 + operatornamecsch^2(u))du\
                      &= u - coth(u)\
                      &= operatornamearctanh(x) - frac1x + C
                      endalign
                      $$






                      share|cite|improve this answer















                      $x = tanh(u)$ then $dx = operatornamesech^2(u) du$
                      $$
                      beginalign
                      int fracdxx^2(1-x^2)
                      &= int fracoperatornamesech^2(u) dutanh^2(u)(1-tanh^2(u))\
                      &= int fracoperatornamesech^2(u) dutanh^2(u)(operatornamesech^2(u))\
                      &= int fracdutanh^2(u)\
                      &= int fraccosh^2(u)sinh^2(u) du\
                      &= int (1 + operatornamecsch^2(u))du\
                      &= u - coth(u)\
                      &= operatornamearctanh(x) - frac1x + C
                      endalign
                      $$







                      share|cite|improve this answer















                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Aug 2 at 12:43









                      user 108128

                      18.8k41544




                      18.8k41544











                      answered Aug 2 at 12:27









                      Cato

                      1,09038




                      1,09038











                      • I've edited yours, if you don't want, so do roll back.
                        – user 108128
                        Aug 2 at 12:44










                      • I think the original questioners partial fraction method was better - apart from missing the factorization of $1 - x^2$ - it was probably easier. I once integrated $frac11 - x^2$ the same sort of way and handed it in to someone
                        – Cato
                        Aug 2 at 13:00

















                      • I've edited yours, if you don't want, so do roll back.
                        – user 108128
                        Aug 2 at 12:44










                      • I think the original questioners partial fraction method was better - apart from missing the factorization of $1 - x^2$ - it was probably easier. I once integrated $frac11 - x^2$ the same sort of way and handed it in to someone
                        – Cato
                        Aug 2 at 13:00
















                      I've edited yours, if you don't want, so do roll back.
                      – user 108128
                      Aug 2 at 12:44




                      I've edited yours, if you don't want, so do roll back.
                      – user 108128
                      Aug 2 at 12:44












                      I think the original questioners partial fraction method was better - apart from missing the factorization of $1 - x^2$ - it was probably easier. I once integrated $frac11 - x^2$ the same sort of way and handed it in to someone
                      – Cato
                      Aug 2 at 13:00





                      I think the original questioners partial fraction method was better - apart from missing the factorization of $1 - x^2$ - it was probably easier. I once integrated $frac11 - x^2$ the same sort of way and handed it in to someone
                      – Cato
                      Aug 2 at 13:00











                      up vote
                      1
                      down vote













                      Note that, if $f(x)=arctan(xi)$, then
                      $$
                      f'(x)=ifrac11+(xi)^2=fraci1-x^2
                      $$
                      that's not the required derivative; you are missing $1/i=-i$, so you could write
                      $$
                      -frac1x-iarctan(xi)
                      $$



                      I'd prefer to avoid complex functions. With partial fractions, we set
                      $$
                      frac1x^2(1-x)^2=
                      fracax+fracbx^2+fraccx-1+fracdx+1
                      $$
                      Removing the denominators leads to
                      $$
                      ax(x^2-1)+b(x^2-1)+cx^2(x+1)+dx^2(x-1)=-1
                      $$
                      With $x=0$, we get $-b=-1$; with $x=1$, we get $2c=-1$; with $x=-1$ we get $-2d=-1$. It remains to find $a$, but this is easily seen to be $0$.



                      Therefore
                      $$
                      intfrac1x^2(1-x^2),dx=
                      intleft(
                      frac1x^2-frac12frac1x-1+frac12frac1x+1
                      right),dx=-frac1x+frac12logleft|fracx+1x-1right|+c
                      $$






                      share|cite|improve this answer























                      • +1. Good illustration. end of solution is $frac12$ behind $log$.
                        – user 108128
                        Aug 2 at 13:38











                      • @user108128 Thanks for noting
                        – egreg
                        Aug 2 at 14:14














                      up vote
                      1
                      down vote













                      Note that, if $f(x)=arctan(xi)$, then
                      $$
                      f'(x)=ifrac11+(xi)^2=fraci1-x^2
                      $$
                      that's not the required derivative; you are missing $1/i=-i$, so you could write
                      $$
                      -frac1x-iarctan(xi)
                      $$



                      I'd prefer to avoid complex functions. With partial fractions, we set
                      $$
                      frac1x^2(1-x)^2=
                      fracax+fracbx^2+fraccx-1+fracdx+1
                      $$
                      Removing the denominators leads to
                      $$
                      ax(x^2-1)+b(x^2-1)+cx^2(x+1)+dx^2(x-1)=-1
                      $$
                      With $x=0$, we get $-b=-1$; with $x=1$, we get $2c=-1$; with $x=-1$ we get $-2d=-1$. It remains to find $a$, but this is easily seen to be $0$.



                      Therefore
                      $$
                      intfrac1x^2(1-x^2),dx=
                      intleft(
                      frac1x^2-frac12frac1x-1+frac12frac1x+1
                      right),dx=-frac1x+frac12logleft|fracx+1x-1right|+c
                      $$






                      share|cite|improve this answer























                      • +1. Good illustration. end of solution is $frac12$ behind $log$.
                        – user 108128
                        Aug 2 at 13:38











                      • @user108128 Thanks for noting
                        – egreg
                        Aug 2 at 14:14












                      up vote
                      1
                      down vote










                      up vote
                      1
                      down vote









                      Note that, if $f(x)=arctan(xi)$, then
                      $$
                      f'(x)=ifrac11+(xi)^2=fraci1-x^2
                      $$
                      that's not the required derivative; you are missing $1/i=-i$, so you could write
                      $$
                      -frac1x-iarctan(xi)
                      $$



                      I'd prefer to avoid complex functions. With partial fractions, we set
                      $$
                      frac1x^2(1-x)^2=
                      fracax+fracbx^2+fraccx-1+fracdx+1
                      $$
                      Removing the denominators leads to
                      $$
                      ax(x^2-1)+b(x^2-1)+cx^2(x+1)+dx^2(x-1)=-1
                      $$
                      With $x=0$, we get $-b=-1$; with $x=1$, we get $2c=-1$; with $x=-1$ we get $-2d=-1$. It remains to find $a$, but this is easily seen to be $0$.



                      Therefore
                      $$
                      intfrac1x^2(1-x^2),dx=
                      intleft(
                      frac1x^2-frac12frac1x-1+frac12frac1x+1
                      right),dx=-frac1x+frac12logleft|fracx+1x-1right|+c
                      $$






                      share|cite|improve this answer















                      Note that, if $f(x)=arctan(xi)$, then
                      $$
                      f'(x)=ifrac11+(xi)^2=fraci1-x^2
                      $$
                      that's not the required derivative; you are missing $1/i=-i$, so you could write
                      $$
                      -frac1x-iarctan(xi)
                      $$



                      I'd prefer to avoid complex functions. With partial fractions, we set
                      $$
                      frac1x^2(1-x)^2=
                      fracax+fracbx^2+fraccx-1+fracdx+1
                      $$
                      Removing the denominators leads to
                      $$
                      ax(x^2-1)+b(x^2-1)+cx^2(x+1)+dx^2(x-1)=-1
                      $$
                      With $x=0$, we get $-b=-1$; with $x=1$, we get $2c=-1$; with $x=-1$ we get $-2d=-1$. It remains to find $a$, but this is easily seen to be $0$.



                      Therefore
                      $$
                      intfrac1x^2(1-x^2),dx=
                      intleft(
                      frac1x^2-frac12frac1x-1+frac12frac1x+1
                      right),dx=-frac1x+frac12logleft|fracx+1x-1right|+c
                      $$







                      share|cite|improve this answer















                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Aug 2 at 14:14


























                      answered Aug 2 at 13:19









                      egreg

                      164k1180187




                      164k1180187











                      • +1. Good illustration. end of solution is $frac12$ behind $log$.
                        – user 108128
                        Aug 2 at 13:38











                      • @user108128 Thanks for noting
                        – egreg
                        Aug 2 at 14:14
















                      • +1. Good illustration. end of solution is $frac12$ behind $log$.
                        – user 108128
                        Aug 2 at 13:38











                      • @user108128 Thanks for noting
                        – egreg
                        Aug 2 at 14:14















                      +1. Good illustration. end of solution is $frac12$ behind $log$.
                      – user 108128
                      Aug 2 at 13:38





                      +1. Good illustration. end of solution is $frac12$ behind $log$.
                      – user 108128
                      Aug 2 at 13:38













                      @user108128 Thanks for noting
                      – egreg
                      Aug 2 at 14:14




                      @user108128 Thanks for noting
                      – egreg
                      Aug 2 at 14:14












                       

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