How to quickly solve partial fractions equation?












2












$begingroup$


Often I am dealing with an integral of let's say:



$$intfrac{dt}{(t-2)(t+3)}$$



or



$$int frac{dt}{t(t-4)}$$



or to make this a more general case in which I am interested the most:



$$int frac{dt}{(t+alpha)(t+beta)} quad quad alpha, beta in mathbb{R}$$



Basically any integral with decomposed polynomial of second degree in the denominator. Obviously this leads to sum of two natural logarithms.



What I do is write down partial fractions equation and then solve system of linear equations (2 variables here):



$$frac{1}{(t+alpha)(t+beta)} = frac{A}{t+alpha} + frac{B}{t+beta}$$



After solving this I end up with some $A, B$ coefficients and I can solve the integral.



Is there faster way to find $A, B$ ? Some algorithm or anything that I can follow and would always work for such case? Surely, solving that does not take much time but I just wonder if it could be done even faster.



(bonus question: what if there were more variables, like 3 variables?)



I would greatly appreciate all feedback as it could help me save countless number of minutes in the future.






calculus integration indefinite-integrals quadratics partial-fractions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Don't you mean find $A,B$? $alpha,beta$ are usually given.
    $endgroup$
    – Eevee Trainer
    3 hours ago










  • $begingroup$
    Yes. Fixed. Thank you!
    $endgroup$
    – weno
    3 hours ago










  • $begingroup$
    Also when I think about it, I wonder if there's even a general method (beyond the one you referred to). I can see very easily a general solution for when (i) the numerator is $1$ (ii) $alpha ne beta$ (iii) we only have a product of two factors in the denominator, but that's such a narrow set of circumstances that I wonder if it's worth it. Maybe someone has a more general form though.
    $endgroup$
    – Eevee Trainer
    3 hours ago
















2












$begingroup$


Often I am dealing with an integral of let's say:



$$intfrac{dt}{(t-2)(t+3)}$$



or



$$int frac{dt}{t(t-4)}$$



or to make this a more general case in which I am interested the most:



$$int frac{dt}{(t+alpha)(t+beta)} quad quad alpha, beta in mathbb{R}$$



Basically any integral with decomposed polynomial of second degree in the denominator. Obviously this leads to sum of two natural logarithms.



What I do is write down partial fractions equation and then solve system of linear equations (2 variables here):



$$frac{1}{(t+alpha)(t+beta)} = frac{A}{t+alpha} + frac{B}{t+beta}$$



After solving this I end up with some $A, B$ coefficients and I can solve the integral.



Is there faster way to find $A, B$ ? Some algorithm or anything that I can follow and would always work for such case? Surely, solving that does not take much time but I just wonder if it could be done even faster.



(bonus question: what if there were more variables, like 3 variables?)



I would greatly appreciate all feedback as it could help me save countless number of minutes in the future.






calculus integration indefinite-integrals quadratics partial-fractions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Don't you mean find $A,B$? $alpha,beta$ are usually given.
    $endgroup$
    – Eevee Trainer
    3 hours ago










  • $begingroup$
    Yes. Fixed. Thank you!
    $endgroup$
    – weno
    3 hours ago










  • $begingroup$
    Also when I think about it, I wonder if there's even a general method (beyond the one you referred to). I can see very easily a general solution for when (i) the numerator is $1$ (ii) $alpha ne beta$ (iii) we only have a product of two factors in the denominator, but that's such a narrow set of circumstances that I wonder if it's worth it. Maybe someone has a more general form though.
    $endgroup$
    – Eevee Trainer
    3 hours ago














2












2








2





$begingroup$


Often I am dealing with an integral of let's say:



$$intfrac{dt}{(t-2)(t+3)}$$



or



$$int frac{dt}{t(t-4)}$$



or to make this a more general case in which I am interested the most:



$$int frac{dt}{(t+alpha)(t+beta)} quad quad alpha, beta in mathbb{R}$$



Basically any integral with decomposed polynomial of second degree in the denominator. Obviously this leads to sum of two natural logarithms.



What I do is write down partial fractions equation and then solve system of linear equations (2 variables here):



$$frac{1}{(t+alpha)(t+beta)} = frac{A}{t+alpha} + frac{B}{t+beta}$$



After solving this I end up with some $A, B$ coefficients and I can solve the integral.



Is there faster way to find $A, B$ ? Some algorithm or anything that I can follow and would always work for such case? Surely, solving that does not take much time but I just wonder if it could be done even faster.



(bonus question: what if there were more variables, like 3 variables?)



I would greatly appreciate all feedback as it could help me save countless number of minutes in the future.






calculus integration indefinite-integrals quadratics partial-fractions







share|cite|improve this question











$endgroup$




Often I am dealing with an integral of let's say:



$$intfrac{dt}{(t-2)(t+3)}$$



or



$$int frac{dt}{t(t-4)}$$



or to make this a more general case in which I am interested the most:



$$int frac{dt}{(t+alpha)(t+beta)} quad quad alpha, beta in mathbb{R}$$



Basically any integral with decomposed polynomial of second degree in the denominator. Obviously this leads to sum of two natural logarithms.



What I do is write down partial fractions equation and then solve system of linear equations (2 variables here):



$$frac{1}{(t+alpha)(t+beta)} = frac{A}{t+alpha} + frac{B}{t+beta}$$



After solving this I end up with some $A, B$ coefficients and I can solve the integral.



Is there faster way to find $A, B$ ? Some algorithm or anything that I can follow and would always work for such case? Surely, solving that does not take much time but I just wonder if it could be done even faster.



(bonus question: what if there were more variables, like 3 variables?)



I would greatly appreciate all feedback as it could help me save countless number of minutes in the future.






calculus integration indefinite-integrals quadratics partial-fractions




calculus integration indefinite-integrals quadratics partial-fractions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago







weno

















asked 3 hours ago









wenoweno

42311




42311








  • 1




    $begingroup$
    Don't you mean find $A,B$? $alpha,beta$ are usually given.
    $endgroup$
    – Eevee Trainer
    3 hours ago










  • $begingroup$
    Yes. Fixed. Thank you!
    $endgroup$
    – weno
    3 hours ago










  • $begingroup$
    Also when I think about it, I wonder if there's even a general method (beyond the one you referred to). I can see very easily a general solution for when (i) the numerator is $1$ (ii) $alpha ne beta$ (iii) we only have a product of two factors in the denominator, but that's such a narrow set of circumstances that I wonder if it's worth it. Maybe someone has a more general form though.
    $endgroup$
    – Eevee Trainer
    3 hours ago














  • 1




    $begingroup$
    Don't you mean find $A,B$? $alpha,beta$ are usually given.
    $endgroup$
    – Eevee Trainer
    3 hours ago










  • $begingroup$
    Yes. Fixed. Thank you!
    $endgroup$
    – weno
    3 hours ago










  • $begingroup$
    Also when I think about it, I wonder if there's even a general method (beyond the one you referred to). I can see very easily a general solution for when (i) the numerator is $1$ (ii) $alpha ne beta$ (iii) we only have a product of two factors in the denominator, but that's such a narrow set of circumstances that I wonder if it's worth it. Maybe someone has a more general form though.
    $endgroup$
    – Eevee Trainer
    3 hours ago








1




1




$begingroup$
Don't you mean find $A,B$? $alpha,beta$ are usually given.
$endgroup$
– Eevee Trainer
3 hours ago




$begingroup$
Don't you mean find $A,B$? $alpha,beta$ are usually given.
$endgroup$
– Eevee Trainer
3 hours ago












$begingroup$
Yes. Fixed. Thank you!
$endgroup$
– weno
3 hours ago




$begingroup$
Yes. Fixed. Thank you!
$endgroup$
– weno
3 hours ago












$begingroup$
Also when I think about it, I wonder if there's even a general method (beyond the one you referred to). I can see very easily a general solution for when (i) the numerator is $1$ (ii) $alpha ne beta$ (iii) we only have a product of two factors in the denominator, but that's such a narrow set of circumstances that I wonder if it's worth it. Maybe someone has a more general form though.
$endgroup$
– Eevee Trainer
3 hours ago




$begingroup$
Also when I think about it, I wonder if there's even a general method (beyond the one you referred to). I can see very easily a general solution for when (i) the numerator is $1$ (ii) $alpha ne beta$ (iii) we only have a product of two factors in the denominator, but that's such a narrow set of circumstances that I wonder if it's worth it. Maybe someone has a more general form though.
$endgroup$
– Eevee Trainer
3 hours ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

Unless I am misunderstanding the question, by "root rotation" (when $beta neq alpha$):



$$frac{1}{(t + alpha)(t + beta)} = frac{A}{t + alpha} + frac{B}{t + beta}$$



$$1 = A(t + beta) + B(t + alpha)$$



Evaluating $beta$ for $t$:



$$1 = B(alpha - beta)$$



$$B = frac{1}{alpha - beta}$$



Similarly, for $A$, sub in $-alpha$:



$$1 = A(beta - alpha)$$



$$A = frac{1}{beta - alpha}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I'll be coming back to this post. This is what I was looking for.
    $endgroup$
    – weno
    2 hours ago










  • $begingroup$
    Unless that takes you too much time, would you be able to derive formulas for $A, B, C$, please? Function of form $frac{1}{(t+alpha)(t+beta)(t+gamma)}$ and $alpha neq beta neq gamma$
    $endgroup$
    – weno
    2 hours ago





















1












$begingroup$

If your fraction is in form of $$ frac {1}{(t-alpha_1)(t-alpha_2)(t-alpha_3)...(t-alpha_k)}$$ where the $alpha s$ are different, there is an easy way to find the partial fraction decomposition.



Let $t=alpha_i$ and cover $(t-alpha_i)$ while evaluating the above fraction to get your $A_i$



For example $$ frac{1}{(t-1)(t-3)(t+4)} = frac {A_1}{(t-1)} + frac{ A_2}{(t-3)} +frac { A_3}{(t+4)}$$
Where $$A_1 = frac {1}{(1-3)(1+4)}=frac{-1}{10}$$
$$A_2 = frac {1}{(3-1)(3+4)} = frac{1}{14}$$



$$A_3=frac {1}{(-4-1)(-4-3)}=frac{1}{35}$$






share|cite|improve this answer









$endgroup$














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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Unless I am misunderstanding the question, by "root rotation" (when $beta neq alpha$):



    $$frac{1}{(t + alpha)(t + beta)} = frac{A}{t + alpha} + frac{B}{t + beta}$$



    $$1 = A(t + beta) + B(t + alpha)$$



    Evaluating $beta$ for $t$:



    $$1 = B(alpha - beta)$$



    $$B = frac{1}{alpha - beta}$$



    Similarly, for $A$, sub in $-alpha$:



    $$1 = A(beta - alpha)$$



    $$A = frac{1}{beta - alpha}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I'll be coming back to this post. This is what I was looking for.
      $endgroup$
      – weno
      2 hours ago










    • $begingroup$
      Unless that takes you too much time, would you be able to derive formulas for $A, B, C$, please? Function of form $frac{1}{(t+alpha)(t+beta)(t+gamma)}$ and $alpha neq beta neq gamma$
      $endgroup$
      – weno
      2 hours ago


















    2












    $begingroup$

    Unless I am misunderstanding the question, by "root rotation" (when $beta neq alpha$):



    $$frac{1}{(t + alpha)(t + beta)} = frac{A}{t + alpha} + frac{B}{t + beta}$$



    $$1 = A(t + beta) + B(t + alpha)$$



    Evaluating $beta$ for $t$:



    $$1 = B(alpha - beta)$$



    $$B = frac{1}{alpha - beta}$$



    Similarly, for $A$, sub in $-alpha$:



    $$1 = A(beta - alpha)$$



    $$A = frac{1}{beta - alpha}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I'll be coming back to this post. This is what I was looking for.
      $endgroup$
      – weno
      2 hours ago










    • $begingroup$
      Unless that takes you too much time, would you be able to derive formulas for $A, B, C$, please? Function of form $frac{1}{(t+alpha)(t+beta)(t+gamma)}$ and $alpha neq beta neq gamma$
      $endgroup$
      – weno
      2 hours ago
















    2












    2








    2





    $begingroup$

    Unless I am misunderstanding the question, by "root rotation" (when $beta neq alpha$):



    $$frac{1}{(t + alpha)(t + beta)} = frac{A}{t + alpha} + frac{B}{t + beta}$$



    $$1 = A(t + beta) + B(t + alpha)$$



    Evaluating $beta$ for $t$:



    $$1 = B(alpha - beta)$$



    $$B = frac{1}{alpha - beta}$$



    Similarly, for $A$, sub in $-alpha$:



    $$1 = A(beta - alpha)$$



    $$A = frac{1}{beta - alpha}$$






    share|cite|improve this answer









    $endgroup$



    Unless I am misunderstanding the question, by "root rotation" (when $beta neq alpha$):



    $$frac{1}{(t + alpha)(t + beta)} = frac{A}{t + alpha} + frac{B}{t + beta}$$



    $$1 = A(t + beta) + B(t + alpha)$$



    Evaluating $beta$ for $t$:



    $$1 = B(alpha - beta)$$



    $$B = frac{1}{alpha - beta}$$



    Similarly, for $A$, sub in $-alpha$:



    $$1 = A(beta - alpha)$$



    $$A = frac{1}{beta - alpha}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 3 hours ago









    DairDair

    1,95711124




    1,95711124












    • $begingroup$
      I'll be coming back to this post. This is what I was looking for.
      $endgroup$
      – weno
      2 hours ago










    • $begingroup$
      Unless that takes you too much time, would you be able to derive formulas for $A, B, C$, please? Function of form $frac{1}{(t+alpha)(t+beta)(t+gamma)}$ and $alpha neq beta neq gamma$
      $endgroup$
      – weno
      2 hours ago




















    • $begingroup$
      I'll be coming back to this post. This is what I was looking for.
      $endgroup$
      – weno
      2 hours ago










    • $begingroup$
      Unless that takes you too much time, would you be able to derive formulas for $A, B, C$, please? Function of form $frac{1}{(t+alpha)(t+beta)(t+gamma)}$ and $alpha neq beta neq gamma$
      $endgroup$
      – weno
      2 hours ago


















    $begingroup$
    I'll be coming back to this post. This is what I was looking for.
    $endgroup$
    – weno
    2 hours ago




    $begingroup$
    I'll be coming back to this post. This is what I was looking for.
    $endgroup$
    – weno
    2 hours ago












    $begingroup$
    Unless that takes you too much time, would you be able to derive formulas for $A, B, C$, please? Function of form $frac{1}{(t+alpha)(t+beta)(t+gamma)}$ and $alpha neq beta neq gamma$
    $endgroup$
    – weno
    2 hours ago






    $begingroup$
    Unless that takes you too much time, would you be able to derive formulas for $A, B, C$, please? Function of form $frac{1}{(t+alpha)(t+beta)(t+gamma)}$ and $alpha neq beta neq gamma$
    $endgroup$
    – weno
    2 hours ago













    1












    $begingroup$

    If your fraction is in form of $$ frac {1}{(t-alpha_1)(t-alpha_2)(t-alpha_3)...(t-alpha_k)}$$ where the $alpha s$ are different, there is an easy way to find the partial fraction decomposition.



    Let $t=alpha_i$ and cover $(t-alpha_i)$ while evaluating the above fraction to get your $A_i$



    For example $$ frac{1}{(t-1)(t-3)(t+4)} = frac {A_1}{(t-1)} + frac{ A_2}{(t-3)} +frac { A_3}{(t+4)}$$
    Where $$A_1 = frac {1}{(1-3)(1+4)}=frac{-1}{10}$$
    $$A_2 = frac {1}{(3-1)(3+4)} = frac{1}{14}$$



    $$A_3=frac {1}{(-4-1)(-4-3)}=frac{1}{35}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      If your fraction is in form of $$ frac {1}{(t-alpha_1)(t-alpha_2)(t-alpha_3)...(t-alpha_k)}$$ where the $alpha s$ are different, there is an easy way to find the partial fraction decomposition.



      Let $t=alpha_i$ and cover $(t-alpha_i)$ while evaluating the above fraction to get your $A_i$



      For example $$ frac{1}{(t-1)(t-3)(t+4)} = frac {A_1}{(t-1)} + frac{ A_2}{(t-3)} +frac { A_3}{(t+4)}$$
      Where $$A_1 = frac {1}{(1-3)(1+4)}=frac{-1}{10}$$
      $$A_2 = frac {1}{(3-1)(3+4)} = frac{1}{14}$$



      $$A_3=frac {1}{(-4-1)(-4-3)}=frac{1}{35}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        If your fraction is in form of $$ frac {1}{(t-alpha_1)(t-alpha_2)(t-alpha_3)...(t-alpha_k)}$$ where the $alpha s$ are different, there is an easy way to find the partial fraction decomposition.



        Let $t=alpha_i$ and cover $(t-alpha_i)$ while evaluating the above fraction to get your $A_i$



        For example $$ frac{1}{(t-1)(t-3)(t+4)} = frac {A_1}{(t-1)} + frac{ A_2}{(t-3)} +frac { A_3}{(t+4)}$$
        Where $$A_1 = frac {1}{(1-3)(1+4)}=frac{-1}{10}$$
        $$A_2 = frac {1}{(3-1)(3+4)} = frac{1}{14}$$



        $$A_3=frac {1}{(-4-1)(-4-3)}=frac{1}{35}$$






        share|cite|improve this answer









        $endgroup$



        If your fraction is in form of $$ frac {1}{(t-alpha_1)(t-alpha_2)(t-alpha_3)...(t-alpha_k)}$$ where the $alpha s$ are different, there is an easy way to find the partial fraction decomposition.



        Let $t=alpha_i$ and cover $(t-alpha_i)$ while evaluating the above fraction to get your $A_i$



        For example $$ frac{1}{(t-1)(t-3)(t+4)} = frac {A_1}{(t-1)} + frac{ A_2}{(t-3)} +frac { A_3}{(t+4)}$$
        Where $$A_1 = frac {1}{(1-3)(1+4)}=frac{-1}{10}$$
        $$A_2 = frac {1}{(3-1)(3+4)} = frac{1}{14}$$



        $$A_3=frac {1}{(-4-1)(-4-3)}=frac{1}{35}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        Mohammad Riazi-KermaniMohammad Riazi-Kermani

        41.6k42061




        41.6k42061






























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            Hulsita

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            Hulsita Hulsita de la localitat tipus Fórmula química Fe 2+ 2 Fe 3+ O 2 BO 3 Epònim Alfred Hulse Brooks Localitat tipus mont Brooks, península de Seward, Nome Borough, Alaska, Estats Units Classificació Categoria borats Nickel-Strunz 10a ed. 6.AB.45 Nickel-Strunz 9a ed. 6.AB.45 Nickel-Strunz 8a ed. V/G.03 Dana 24.2.3.1 Heys 9.6.4 Propietats Sistema cristal·lí monoclínic Estructura cristal·lina a = 10,68Å; b = 3,09Å; c = 5,43Å; β = 94,15° Color negre Macles en [001]; un membre de la macla es troba girat 120° en relació a l'altre Exfoliació bona en {110} Duresa 3 Lluïssor vítria, submetàl·lica Densitat 4,28 g/cm 3 (mesurada); Més informació Estatus IMA mineral heretat (G) i mineral heretat (G) Any d'aprovació 1908 Referències [1] La hulsita és un mineral de la classe dels borats, que pertany al grup de la pinakiolita. Rep el seu nom en honor del nord-americà
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