Tedious fraction decomposition integral $intfrac{1}{(x^2-1)^2} , dx$












1














I'm having this integral to resolve using fraction decomposition:



$$intfrac{1}{(x^2-1)^2} , dx$$



This gives me the following:



$$frac{A}{x+1} + frac{B}{(x+1)^2} + frac{C}{x-1} + frac{D}{(x-1)^2} = frac{1}{(x^2-1)^2}$$



Which results in a 4 variables system of equation...



Is there a quick way to resolve this that I'm missing out?






integration indefinite-integrals partial-fractions







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  • 2




    The right side should be $frac{1}{(x^2-1)^2}$, not $1$. A method that speeds a lot of these up is Heaviside's method: clear denominators, and plug in the roots of your polynomial in one at a time. That gives a few relations that you may use to speed up what you're trying to do.
    – Barry Smith
    Feb 23 '16 at 2:34










  • Another method would be substituting $x=sec theta$, then breaking the transformed integral down into sines and cosines, and then performing a $u$-sub. But since it said to specifically use partial-fraction decomposition...
    – Barry Smith
    Feb 23 '16 at 2:38






  • 1




    Or: type "partial fraction decomposition for 1/(1-x^2)^2" into wolframalpha.com. Once you have the decomposition, you can check that it is correct by making a common denominator. This is a perfectly valid way to produce a decomposition, although if this is for homework, it may not be one that your grader accepts.
    – Barry Smith
    Feb 23 '16 at 2:43
















1














I'm having this integral to resolve using fraction decomposition:



$$intfrac{1}{(x^2-1)^2} , dx$$



This gives me the following:



$$frac{A}{x+1} + frac{B}{(x+1)^2} + frac{C}{x-1} + frac{D}{(x-1)^2} = frac{1}{(x^2-1)^2}$$



Which results in a 4 variables system of equation...



Is there a quick way to resolve this that I'm missing out?






integration indefinite-integrals partial-fractions







share|cite|improve this question




















  • 2




    The right side should be $frac{1}{(x^2-1)^2}$, not $1$. A method that speeds a lot of these up is Heaviside's method: clear denominators, and plug in the roots of your polynomial in one at a time. That gives a few relations that you may use to speed up what you're trying to do.
    – Barry Smith
    Feb 23 '16 at 2:34










  • Another method would be substituting $x=sec theta$, then breaking the transformed integral down into sines and cosines, and then performing a $u$-sub. But since it said to specifically use partial-fraction decomposition...
    – Barry Smith
    Feb 23 '16 at 2:38






  • 1




    Or: type "partial fraction decomposition for 1/(1-x^2)^2" into wolframalpha.com. Once you have the decomposition, you can check that it is correct by making a common denominator. This is a perfectly valid way to produce a decomposition, although if this is for homework, it may not be one that your grader accepts.
    – Barry Smith
    Feb 23 '16 at 2:43














1












1








1







I'm having this integral to resolve using fraction decomposition:



$$intfrac{1}{(x^2-1)^2} , dx$$



This gives me the following:



$$frac{A}{x+1} + frac{B}{(x+1)^2} + frac{C}{x-1} + frac{D}{(x-1)^2} = frac{1}{(x^2-1)^2}$$



Which results in a 4 variables system of equation...



Is there a quick way to resolve this that I'm missing out?






integration indefinite-integrals partial-fractions







share|cite|improve this question















I'm having this integral to resolve using fraction decomposition:



$$intfrac{1}{(x^2-1)^2} , dx$$



This gives me the following:



$$frac{A}{x+1} + frac{B}{(x+1)^2} + frac{C}{x-1} + frac{D}{(x-1)^2} = frac{1}{(x^2-1)^2}$$



Which results in a 4 variables system of equation...



Is there a quick way to resolve this that I'm missing out?






integration indefinite-integrals partial-fractions




integration indefinite-integrals partial-fractions






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share|cite|improve this question













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edited Nov 30 at 19:03









Martin Sleziak

44.7k7115270




44.7k7115270










asked Feb 23 '16 at 2:30









Machinegon

1651212




1651212








  • 2




    The right side should be $frac{1}{(x^2-1)^2}$, not $1$. A method that speeds a lot of these up is Heaviside's method: clear denominators, and plug in the roots of your polynomial in one at a time. That gives a few relations that you may use to speed up what you're trying to do.
    – Barry Smith
    Feb 23 '16 at 2:34










  • Another method would be substituting $x=sec theta$, then breaking the transformed integral down into sines and cosines, and then performing a $u$-sub. But since it said to specifically use partial-fraction decomposition...
    – Barry Smith
    Feb 23 '16 at 2:38






  • 1




    Or: type "partial fraction decomposition for 1/(1-x^2)^2" into wolframalpha.com. Once you have the decomposition, you can check that it is correct by making a common denominator. This is a perfectly valid way to produce a decomposition, although if this is for homework, it may not be one that your grader accepts.
    – Barry Smith
    Feb 23 '16 at 2:43














  • 2




    The right side should be $frac{1}{(x^2-1)^2}$, not $1$. A method that speeds a lot of these up is Heaviside's method: clear denominators, and plug in the roots of your polynomial in one at a time. That gives a few relations that you may use to speed up what you're trying to do.
    – Barry Smith
    Feb 23 '16 at 2:34










  • Another method would be substituting $x=sec theta$, then breaking the transformed integral down into sines and cosines, and then performing a $u$-sub. But since it said to specifically use partial-fraction decomposition...
    – Barry Smith
    Feb 23 '16 at 2:38






  • 1




    Or: type "partial fraction decomposition for 1/(1-x^2)^2" into wolframalpha.com. Once you have the decomposition, you can check that it is correct by making a common denominator. This is a perfectly valid way to produce a decomposition, although if this is for homework, it may not be one that your grader accepts.
    – Barry Smith
    Feb 23 '16 at 2:43








2




2




The right side should be $frac{1}{(x^2-1)^2}$, not $1$. A method that speeds a lot of these up is Heaviside's method: clear denominators, and plug in the roots of your polynomial in one at a time. That gives a few relations that you may use to speed up what you're trying to do.
– Barry Smith
Feb 23 '16 at 2:34




The right side should be $frac{1}{(x^2-1)^2}$, not $1$. A method that speeds a lot of these up is Heaviside's method: clear denominators, and plug in the roots of your polynomial in one at a time. That gives a few relations that you may use to speed up what you're trying to do.
– Barry Smith
Feb 23 '16 at 2:34












Another method would be substituting $x=sec theta$, then breaking the transformed integral down into sines and cosines, and then performing a $u$-sub. But since it said to specifically use partial-fraction decomposition...
– Barry Smith
Feb 23 '16 at 2:38




Another method would be substituting $x=sec theta$, then breaking the transformed integral down into sines and cosines, and then performing a $u$-sub. But since it said to specifically use partial-fraction decomposition...
– Barry Smith
Feb 23 '16 at 2:38




1




1




Or: type "partial fraction decomposition for 1/(1-x^2)^2" into wolframalpha.com. Once you have the decomposition, you can check that it is correct by making a common denominator. This is a perfectly valid way to produce a decomposition, although if this is for homework, it may not be one that your grader accepts.
– Barry Smith
Feb 23 '16 at 2:43




Or: type "partial fraction decomposition for 1/(1-x^2)^2" into wolframalpha.com. Once you have the decomposition, you can check that it is correct by making a common denominator. This is a perfectly valid way to produce a decomposition, although if this is for homework, it may not be one that your grader accepts.
– Barry Smith
Feb 23 '16 at 2:43










2 Answers
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3














You can work out the partial fraction decomposition of the integrand by repeat application of the identify:
$$frac{1}{(x-a)(x-b)} = frac{1}{(b-a)}left[frac{1}{x-a} - frac{1}{x-b}right]$$
This is especially useful when $a,b$ are small integers.
$$begin{align}frac{1}{(x^2-1)^2}
&= left[frac{1}{(x-1)(x+1)}right]^2 =
frac14left[frac{1}{x-1} - frac{1}{x+1}right]^2\
&= frac14left[frac{1}{(x-1)^2} - frac{2}{(x-1)(x+1)} + frac{1}{(x+1)}^2right]\
&= frac14left[frac{1}{(x-1)^2} - left(frac{1}{x-1} - frac{1}{x+1}right) + frac{1}{(x+1)^2}right]end{align}$$






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    0














    Alternatively start from $frac{1}{(x^2-a)^2} = frac{partial}{partial a} frac{1}{(x^2-a)}$, then determine $intfrac{1}{(x^2-a)} , dx$ (this is straight forward to put the integrand into partial fractions), differentiate this wrt $a$ then set $a=1$.






    share|cite|improve this answer























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      2 Answers
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      2 Answers
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      3














      You can work out the partial fraction decomposition of the integrand by repeat application of the identify:
      $$frac{1}{(x-a)(x-b)} = frac{1}{(b-a)}left[frac{1}{x-a} - frac{1}{x-b}right]$$
      This is especially useful when $a,b$ are small integers.
      $$begin{align}frac{1}{(x^2-1)^2}
      &= left[frac{1}{(x-1)(x+1)}right]^2 =
      frac14left[frac{1}{x-1} - frac{1}{x+1}right]^2\
      &= frac14left[frac{1}{(x-1)^2} - frac{2}{(x-1)(x+1)} + frac{1}{(x+1)}^2right]\
      &= frac14left[frac{1}{(x-1)^2} - left(frac{1}{x-1} - frac{1}{x+1}right) + frac{1}{(x+1)^2}right]end{align}$$






      share|cite|improve this answer


























        3














        You can work out the partial fraction decomposition of the integrand by repeat application of the identify:
        $$frac{1}{(x-a)(x-b)} = frac{1}{(b-a)}left[frac{1}{x-a} - frac{1}{x-b}right]$$
        This is especially useful when $a,b$ are small integers.
        $$begin{align}frac{1}{(x^2-1)^2}
        &= left[frac{1}{(x-1)(x+1)}right]^2 =
        frac14left[frac{1}{x-1} - frac{1}{x+1}right]^2\
        &= frac14left[frac{1}{(x-1)^2} - frac{2}{(x-1)(x+1)} + frac{1}{(x+1)}^2right]\
        &= frac14left[frac{1}{(x-1)^2} - left(frac{1}{x-1} - frac{1}{x+1}right) + frac{1}{(x+1)^2}right]end{align}$$






        share|cite|improve this answer
























          3












          3








          3






          You can work out the partial fraction decomposition of the integrand by repeat application of the identify:
          $$frac{1}{(x-a)(x-b)} = frac{1}{(b-a)}left[frac{1}{x-a} - frac{1}{x-b}right]$$
          This is especially useful when $a,b$ are small integers.
          $$begin{align}frac{1}{(x^2-1)^2}
          &= left[frac{1}{(x-1)(x+1)}right]^2 =
          frac14left[frac{1}{x-1} - frac{1}{x+1}right]^2\
          &= frac14left[frac{1}{(x-1)^2} - frac{2}{(x-1)(x+1)} + frac{1}{(x+1)}^2right]\
          &= frac14left[frac{1}{(x-1)^2} - left(frac{1}{x-1} - frac{1}{x+1}right) + frac{1}{(x+1)^2}right]end{align}$$






          share|cite|improve this answer












          You can work out the partial fraction decomposition of the integrand by repeat application of the identify:
          $$frac{1}{(x-a)(x-b)} = frac{1}{(b-a)}left[frac{1}{x-a} - frac{1}{x-b}right]$$
          This is especially useful when $a,b$ are small integers.
          $$begin{align}frac{1}{(x^2-1)^2}
          &= left[frac{1}{(x-1)(x+1)}right]^2 =
          frac14left[frac{1}{x-1} - frac{1}{x+1}right]^2\
          &= frac14left[frac{1}{(x-1)^2} - frac{2}{(x-1)(x+1)} + frac{1}{(x+1)}^2right]\
          &= frac14left[frac{1}{(x-1)^2} - left(frac{1}{x-1} - frac{1}{x+1}right) + frac{1}{(x+1)^2}right]end{align}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 23 '16 at 2:45









          achille hui

          95.3k5130256




          95.3k5130256























              0














              Alternatively start from $frac{1}{(x^2-a)^2} = frac{partial}{partial a} frac{1}{(x^2-a)}$, then determine $intfrac{1}{(x^2-a)} , dx$ (this is straight forward to put the integrand into partial fractions), differentiate this wrt $a$ then set $a=1$.






              share|cite|improve this answer




























                0














                Alternatively start from $frac{1}{(x^2-a)^2} = frac{partial}{partial a} frac{1}{(x^2-a)}$, then determine $intfrac{1}{(x^2-a)} , dx$ (this is straight forward to put the integrand into partial fractions), differentiate this wrt $a$ then set $a=1$.






                share|cite|improve this answer


























                  0












                  0








                  0






                  Alternatively start from $frac{1}{(x^2-a)^2} = frac{partial}{partial a} frac{1}{(x^2-a)}$, then determine $intfrac{1}{(x^2-a)} , dx$ (this is straight forward to put the integrand into partial fractions), differentiate this wrt $a$ then set $a=1$.






                  share|cite|improve this answer














                  Alternatively start from $frac{1}{(x^2-a)^2} = frac{partial}{partial a} frac{1}{(x^2-a)}$, then determine $intfrac{1}{(x^2-a)} , dx$ (this is straight forward to put the integrand into partial fractions), differentiate this wrt $a$ then set $a=1$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Feb 23 '16 at 8:38

























                  answered Feb 23 '16 at 8:31









                  jim

                  978614




                  978614






























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