Integrate $displaystyleintfrac{1}{u^2 - 1},du$ without partial fractions?












1












$begingroup$


Is there any way possible that I might integrate
$$ intfrac{1}{u^2-1},du $$ without appealing to partial fraction decomposition?



I am trying to work some interesting $u$-substitution integrals with novice students who do not need to be taught partial fraction decomposition.



Thank you.






calculus integration







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$endgroup$








  • 5




    $begingroup$
    Why don't they need to be taught PFD?
    $endgroup$
    – Andrew Li
    Apr 14 '18 at 16:07






  • 1




    $begingroup$
    consider trigonometry!
    $endgroup$
    – Karn Watcharasupat
    Apr 14 '18 at 16:07






  • 2




    $begingroup$
    Hint: et $u = sec t$.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Apr 14 '18 at 16:07






  • 1




    $begingroup$
    IMO, substitution is harder than partial fraction decomposition, which is just a matter of algebra, not calculus.
    $endgroup$
    – Yves Daoust
    Apr 14 '18 at 16:20








  • 1




    $begingroup$
    $u=cosh x$ probably works just as well as $sec x$, though I suppose hyperbolic functions may not be taught yet
    $endgroup$
    – John Doe
    Apr 14 '18 at 16:25
















1












$begingroup$


Is there any way possible that I might integrate
$$ intfrac{1}{u^2-1},du $$ without appealing to partial fraction decomposition?



I am trying to work some interesting $u$-substitution integrals with novice students who do not need to be taught partial fraction decomposition.



Thank you.






calculus integration







share|cite|improve this question









$endgroup$








  • 5




    $begingroup$
    Why don't they need to be taught PFD?
    $endgroup$
    – Andrew Li
    Apr 14 '18 at 16:07






  • 1




    $begingroup$
    consider trigonometry!
    $endgroup$
    – Karn Watcharasupat
    Apr 14 '18 at 16:07






  • 2




    $begingroup$
    Hint: et $u = sec t$.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Apr 14 '18 at 16:07






  • 1




    $begingroup$
    IMO, substitution is harder than partial fraction decomposition, which is just a matter of algebra, not calculus.
    $endgroup$
    – Yves Daoust
    Apr 14 '18 at 16:20








  • 1




    $begingroup$
    $u=cosh x$ probably works just as well as $sec x$, though I suppose hyperbolic functions may not be taught yet
    $endgroup$
    – John Doe
    Apr 14 '18 at 16:25














1












1








1





$begingroup$


Is there any way possible that I might integrate
$$ intfrac{1}{u^2-1},du $$ without appealing to partial fraction decomposition?



I am trying to work some interesting $u$-substitution integrals with novice students who do not need to be taught partial fraction decomposition.



Thank you.






calculus integration







share|cite|improve this question









$endgroup$




Is there any way possible that I might integrate
$$ intfrac{1}{u^2-1},du $$ without appealing to partial fraction decomposition?



I am trying to work some interesting $u$-substitution integrals with novice students who do not need to be taught partial fraction decomposition.



Thank you.






calculus integration




calculus integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 14 '18 at 16:06









clocktowerclocktower

524418




524418








  • 5




    $begingroup$
    Why don't they need to be taught PFD?
    $endgroup$
    – Andrew Li
    Apr 14 '18 at 16:07






  • 1




    $begingroup$
    consider trigonometry!
    $endgroup$
    – Karn Watcharasupat
    Apr 14 '18 at 16:07






  • 2




    $begingroup$
    Hint: et $u = sec t$.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Apr 14 '18 at 16:07






  • 1




    $begingroup$
    IMO, substitution is harder than partial fraction decomposition, which is just a matter of algebra, not calculus.
    $endgroup$
    – Yves Daoust
    Apr 14 '18 at 16:20








  • 1




    $begingroup$
    $u=cosh x$ probably works just as well as $sec x$, though I suppose hyperbolic functions may not be taught yet
    $endgroup$
    – John Doe
    Apr 14 '18 at 16:25














  • 5




    $begingroup$
    Why don't they need to be taught PFD?
    $endgroup$
    – Andrew Li
    Apr 14 '18 at 16:07






  • 1




    $begingroup$
    consider trigonometry!
    $endgroup$
    – Karn Watcharasupat
    Apr 14 '18 at 16:07






  • 2




    $begingroup$
    Hint: et $u = sec t$.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Apr 14 '18 at 16:07






  • 1




    $begingroup$
    IMO, substitution is harder than partial fraction decomposition, which is just a matter of algebra, not calculus.
    $endgroup$
    – Yves Daoust
    Apr 14 '18 at 16:20








  • 1




    $begingroup$
    $u=cosh x$ probably works just as well as $sec x$, though I suppose hyperbolic functions may not be taught yet
    $endgroup$
    – John Doe
    Apr 14 '18 at 16:25








5




5




$begingroup$
Why don't they need to be taught PFD?
$endgroup$
– Andrew Li
Apr 14 '18 at 16:07




$begingroup$
Why don't they need to be taught PFD?
$endgroup$
– Andrew Li
Apr 14 '18 at 16:07




1




1




$begingroup$
consider trigonometry!
$endgroup$
– Karn Watcharasupat
Apr 14 '18 at 16:07




$begingroup$
consider trigonometry!
$endgroup$
– Karn Watcharasupat
Apr 14 '18 at 16:07




2




2




$begingroup$
Hint: et $u = sec t$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Apr 14 '18 at 16:07




$begingroup$
Hint: et $u = sec t$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Apr 14 '18 at 16:07




1




1




$begingroup$
IMO, substitution is harder than partial fraction decomposition, which is just a matter of algebra, not calculus.
$endgroup$
– Yves Daoust
Apr 14 '18 at 16:20






$begingroup$
IMO, substitution is harder than partial fraction decomposition, which is just a matter of algebra, not calculus.
$endgroup$
– Yves Daoust
Apr 14 '18 at 16:20






1




1




$begingroup$
$u=cosh x$ probably works just as well as $sec x$, though I suppose hyperbolic functions may not be taught yet
$endgroup$
– John Doe
Apr 14 '18 at 16:25




$begingroup$
$u=cosh x$ probably works just as well as $sec x$, though I suppose hyperbolic functions may not be taught yet
$endgroup$
– John Doe
Apr 14 '18 at 16:25










5 Answers
5






active

oldest

votes


















3












$begingroup$

This is motivated by the solution, so it works quite nicely. It is an alternative to the $sec$ substitution that has been suggested.$$u=frac{1-x}{1+x}$$ $$frac1{u^2-1}=-frac{4x}{(1+x)^2},,,,,,,,,,,,,du=-frac2{(1+x)^2},dx\$$ so the integral becomes $$intfrac1{2x},dx=frac12ln x$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    The moment you see $$u^2-1$$
    you should think of some trigonometric stuff.



    So if you remember,



    $$tan^2x=sec^2x-1$$



    thus $$u=sec x$$ surely is a candidate (and there are a lot of other variations to play around with!).






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How would you go about integrating $csc x$? This is what this would lead to. If I was given that, I'd probably end up coming back to this and using partial fractions..
      $endgroup$
      – John Doe
      Apr 14 '18 at 16:10








    • 1




      $begingroup$
      @JohnDoe It's just $$-ln(csc x + cot x)+C$$ You can always teach your student how this was derived since it's pretty manageable.
      $endgroup$
      – Karn Watcharasupat
      Apr 14 '18 at 16:12





















    2












    $begingroup$

    Use trigonometric substitution as the integrand contains $u^2-a^2$ where $a=1$. Substitute $u = asec theta$, or just $u = sec theta$.



    $$u = sec thetaquad mathrm du = secthetatantheta,mathrm dtheta$$
    $$int {1over sec^2 theta - 1} secthetatantheta,mathrm dtheta$$



    Which can be integrated by remembering the identity $tan^2 theta + 1 = sec^2 theta$:



    $$int {1over sec^2 theta - 1} secthetatantheta,mathrm dtheta = int {sec theta over tantheta}mathrm dtheta = int csc theta ,mathrm dtheta$$



    And you can find the antiderivative of $csc theta$ via identities, see this. Of course, this would be much easier via partial fraction decomposition.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      We have
      $$
      frac{1}{1-u^2}=frac{1}{(1+u)(1-u)}=frac{1}{(1+u)(2-(1+u))}=frac{1}{(1+u)^2bigl(frac{2}{1+u}-1bigr)}.
      $$
      Thus,
      $$
      intfrac{1}{1-u^2},du=-frac{1}{2}lnbiggl|frac{2}{1+u}-1biggr|+C.
      $$






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        The answers provided already show many different ways this integral can be approached without Partial Fractions. Here is one more. Similar to the trigonometric substitution $u = sec(x)$ we can employ the hyperbolic substitution $u = tanh(x)$



        begin{align}
        int frac{1}{u^2 - 1}:du &= int frac{1}{tanh^2(x) - 1}cdot -operatorname{sech}^2(x):dx \
        &= int frac{1}{operatorname{sech}^2(x)}cdot -operatorname{sech}^2(x):dx \
        &= -x + C = -operatorname{arctanh(u)} + C
        end{align}



        Where $C$ is the constant of integration.






        share|cite|improve this answer











        $endgroup$














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          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          This is motivated by the solution, so it works quite nicely. It is an alternative to the $sec$ substitution that has been suggested.$$u=frac{1-x}{1+x}$$ $$frac1{u^2-1}=-frac{4x}{(1+x)^2},,,,,,,,,,,,,du=-frac2{(1+x)^2},dx\$$ so the integral becomes $$intfrac1{2x},dx=frac12ln x$$






          share|cite|improve this answer









          $endgroup$


















            3












            $begingroup$

            This is motivated by the solution, so it works quite nicely. It is an alternative to the $sec$ substitution that has been suggested.$$u=frac{1-x}{1+x}$$ $$frac1{u^2-1}=-frac{4x}{(1+x)^2},,,,,,,,,,,,,du=-frac2{(1+x)^2},dx\$$ so the integral becomes $$intfrac1{2x},dx=frac12ln x$$






            share|cite|improve this answer









            $endgroup$
















              3












              3








              3





              $begingroup$

              This is motivated by the solution, so it works quite nicely. It is an alternative to the $sec$ substitution that has been suggested.$$u=frac{1-x}{1+x}$$ $$frac1{u^2-1}=-frac{4x}{(1+x)^2},,,,,,,,,,,,,du=-frac2{(1+x)^2},dx\$$ so the integral becomes $$intfrac1{2x},dx=frac12ln x$$






              share|cite|improve this answer









              $endgroup$



              This is motivated by the solution, so it works quite nicely. It is an alternative to the $sec$ substitution that has been suggested.$$u=frac{1-x}{1+x}$$ $$frac1{u^2-1}=-frac{4x}{(1+x)^2},,,,,,,,,,,,,du=-frac2{(1+x)^2},dx\$$ so the integral becomes $$intfrac1{2x},dx=frac12ln x$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Apr 14 '18 at 16:20









              John DoeJohn Doe

              11.9k11339




              11.9k11339























                  2












                  $begingroup$

                  The moment you see $$u^2-1$$
                  you should think of some trigonometric stuff.



                  So if you remember,



                  $$tan^2x=sec^2x-1$$



                  thus $$u=sec x$$ surely is a candidate (and there are a lot of other variations to play around with!).






                  share|cite|improve this answer









                  $endgroup$













                  • $begingroup$
                    How would you go about integrating $csc x$? This is what this would lead to. If I was given that, I'd probably end up coming back to this and using partial fractions..
                    $endgroup$
                    – John Doe
                    Apr 14 '18 at 16:10








                  • 1




                    $begingroup$
                    @JohnDoe It's just $$-ln(csc x + cot x)+C$$ You can always teach your student how this was derived since it's pretty manageable.
                    $endgroup$
                    – Karn Watcharasupat
                    Apr 14 '18 at 16:12


















                  2












                  $begingroup$

                  The moment you see $$u^2-1$$
                  you should think of some trigonometric stuff.



                  So if you remember,



                  $$tan^2x=sec^2x-1$$



                  thus $$u=sec x$$ surely is a candidate (and there are a lot of other variations to play around with!).






                  share|cite|improve this answer









                  $endgroup$













                  • $begingroup$
                    How would you go about integrating $csc x$? This is what this would lead to. If I was given that, I'd probably end up coming back to this and using partial fractions..
                    $endgroup$
                    – John Doe
                    Apr 14 '18 at 16:10








                  • 1




                    $begingroup$
                    @JohnDoe It's just $$-ln(csc x + cot x)+C$$ You can always teach your student how this was derived since it's pretty manageable.
                    $endgroup$
                    – Karn Watcharasupat
                    Apr 14 '18 at 16:12
















                  2












                  2








                  2





                  $begingroup$

                  The moment you see $$u^2-1$$
                  you should think of some trigonometric stuff.



                  So if you remember,



                  $$tan^2x=sec^2x-1$$



                  thus $$u=sec x$$ surely is a candidate (and there are a lot of other variations to play around with!).






                  share|cite|improve this answer









                  $endgroup$



                  The moment you see $$u^2-1$$
                  you should think of some trigonometric stuff.



                  So if you remember,



                  $$tan^2x=sec^2x-1$$



                  thus $$u=sec x$$ surely is a candidate (and there are a lot of other variations to play around with!).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Apr 14 '18 at 16:09









                  Karn WatcharasupatKarn Watcharasupat

                  3,9642526




                  3,9642526












                  • $begingroup$
                    How would you go about integrating $csc x$? This is what this would lead to. If I was given that, I'd probably end up coming back to this and using partial fractions..
                    $endgroup$
                    – John Doe
                    Apr 14 '18 at 16:10








                  • 1




                    $begingroup$
                    @JohnDoe It's just $$-ln(csc x + cot x)+C$$ You can always teach your student how this was derived since it's pretty manageable.
                    $endgroup$
                    – Karn Watcharasupat
                    Apr 14 '18 at 16:12




















                  • $begingroup$
                    How would you go about integrating $csc x$? This is what this would lead to. If I was given that, I'd probably end up coming back to this and using partial fractions..
                    $endgroup$
                    – John Doe
                    Apr 14 '18 at 16:10








                  • 1




                    $begingroup$
                    @JohnDoe It's just $$-ln(csc x + cot x)+C$$ You can always teach your student how this was derived since it's pretty manageable.
                    $endgroup$
                    – Karn Watcharasupat
                    Apr 14 '18 at 16:12


















                  $begingroup$
                  How would you go about integrating $csc x$? This is what this would lead to. If I was given that, I'd probably end up coming back to this and using partial fractions..
                  $endgroup$
                  – John Doe
                  Apr 14 '18 at 16:10






                  $begingroup$
                  How would you go about integrating $csc x$? This is what this would lead to. If I was given that, I'd probably end up coming back to this and using partial fractions..
                  $endgroup$
                  – John Doe
                  Apr 14 '18 at 16:10






                  1




                  1




                  $begingroup$
                  @JohnDoe It's just $$-ln(csc x + cot x)+C$$ You can always teach your student how this was derived since it's pretty manageable.
                  $endgroup$
                  – Karn Watcharasupat
                  Apr 14 '18 at 16:12






                  $begingroup$
                  @JohnDoe It's just $$-ln(csc x + cot x)+C$$ You can always teach your student how this was derived since it's pretty manageable.
                  $endgroup$
                  – Karn Watcharasupat
                  Apr 14 '18 at 16:12













                  2












                  $begingroup$

                  Use trigonometric substitution as the integrand contains $u^2-a^2$ where $a=1$. Substitute $u = asec theta$, or just $u = sec theta$.



                  $$u = sec thetaquad mathrm du = secthetatantheta,mathrm dtheta$$
                  $$int {1over sec^2 theta - 1} secthetatantheta,mathrm dtheta$$



                  Which can be integrated by remembering the identity $tan^2 theta + 1 = sec^2 theta$:



                  $$int {1over sec^2 theta - 1} secthetatantheta,mathrm dtheta = int {sec theta over tantheta}mathrm dtheta = int csc theta ,mathrm dtheta$$



                  And you can find the antiderivative of $csc theta$ via identities, see this. Of course, this would be much easier via partial fraction decomposition.






                  share|cite|improve this answer











                  $endgroup$


















                    2












                    $begingroup$

                    Use trigonometric substitution as the integrand contains $u^2-a^2$ where $a=1$. Substitute $u = asec theta$, or just $u = sec theta$.



                    $$u = sec thetaquad mathrm du = secthetatantheta,mathrm dtheta$$
                    $$int {1over sec^2 theta - 1} secthetatantheta,mathrm dtheta$$



                    Which can be integrated by remembering the identity $tan^2 theta + 1 = sec^2 theta$:



                    $$int {1over sec^2 theta - 1} secthetatantheta,mathrm dtheta = int {sec theta over tantheta}mathrm dtheta = int csc theta ,mathrm dtheta$$



                    And you can find the antiderivative of $csc theta$ via identities, see this. Of course, this would be much easier via partial fraction decomposition.






                    share|cite|improve this answer











                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      Use trigonometric substitution as the integrand contains $u^2-a^2$ where $a=1$. Substitute $u = asec theta$, or just $u = sec theta$.



                      $$u = sec thetaquad mathrm du = secthetatantheta,mathrm dtheta$$
                      $$int {1over sec^2 theta - 1} secthetatantheta,mathrm dtheta$$



                      Which can be integrated by remembering the identity $tan^2 theta + 1 = sec^2 theta$:



                      $$int {1over sec^2 theta - 1} secthetatantheta,mathrm dtheta = int {sec theta over tantheta}mathrm dtheta = int csc theta ,mathrm dtheta$$



                      And you can find the antiderivative of $csc theta$ via identities, see this. Of course, this would be much easier via partial fraction decomposition.






                      share|cite|improve this answer











                      $endgroup$



                      Use trigonometric substitution as the integrand contains $u^2-a^2$ where $a=1$. Substitute $u = asec theta$, or just $u = sec theta$.



                      $$u = sec thetaquad mathrm du = secthetatantheta,mathrm dtheta$$
                      $$int {1over sec^2 theta - 1} secthetatantheta,mathrm dtheta$$



                      Which can be integrated by remembering the identity $tan^2 theta + 1 = sec^2 theta$:



                      $$int {1over sec^2 theta - 1} secthetatantheta,mathrm dtheta = int {sec theta over tantheta}mathrm dtheta = int csc theta ,mathrm dtheta$$



                      And you can find the antiderivative of $csc theta$ via identities, see this. Of course, this would be much easier via partial fraction decomposition.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Apr 14 '18 at 16:15

























                      answered Apr 14 '18 at 16:09









                      Andrew LiAndrew Li

                      4,1621927




                      4,1621927























                          1












                          $begingroup$

                          We have
                          $$
                          frac{1}{1-u^2}=frac{1}{(1+u)(1-u)}=frac{1}{(1+u)(2-(1+u))}=frac{1}{(1+u)^2bigl(frac{2}{1+u}-1bigr)}.
                          $$
                          Thus,
                          $$
                          intfrac{1}{1-u^2},du=-frac{1}{2}lnbiggl|frac{2}{1+u}-1biggr|+C.
                          $$






                          share|cite|improve this answer









                          $endgroup$


















                            1












                            $begingroup$

                            We have
                            $$
                            frac{1}{1-u^2}=frac{1}{(1+u)(1-u)}=frac{1}{(1+u)(2-(1+u))}=frac{1}{(1+u)^2bigl(frac{2}{1+u}-1bigr)}.
                            $$
                            Thus,
                            $$
                            intfrac{1}{1-u^2},du=-frac{1}{2}lnbiggl|frac{2}{1+u}-1biggr|+C.
                            $$






                            share|cite|improve this answer









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              We have
                              $$
                              frac{1}{1-u^2}=frac{1}{(1+u)(1-u)}=frac{1}{(1+u)(2-(1+u))}=frac{1}{(1+u)^2bigl(frac{2}{1+u}-1bigr)}.
                              $$
                              Thus,
                              $$
                              intfrac{1}{1-u^2},du=-frac{1}{2}lnbiggl|frac{2}{1+u}-1biggr|+C.
                              $$






                              share|cite|improve this answer









                              $endgroup$



                              We have
                              $$
                              frac{1}{1-u^2}=frac{1}{(1+u)(1-u)}=frac{1}{(1+u)(2-(1+u))}=frac{1}{(1+u)^2bigl(frac{2}{1+u}-1bigr)}.
                              $$
                              Thus,
                              $$
                              intfrac{1}{1-u^2},du=-frac{1}{2}lnbiggl|frac{2}{1+u}-1biggr|+C.
                              $$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Apr 14 '18 at 17:59









                              mickepmickep

                              18.7k12351




                              18.7k12351























                                  0












                                  $begingroup$

                                  The answers provided already show many different ways this integral can be approached without Partial Fractions. Here is one more. Similar to the trigonometric substitution $u = sec(x)$ we can employ the hyperbolic substitution $u = tanh(x)$



                                  begin{align}
                                  int frac{1}{u^2 - 1}:du &= int frac{1}{tanh^2(x) - 1}cdot -operatorname{sech}^2(x):dx \
                                  &= int frac{1}{operatorname{sech}^2(x)}cdot -operatorname{sech}^2(x):dx \
                                  &= -x + C = -operatorname{arctanh(u)} + C
                                  end{align}



                                  Where $C$ is the constant of integration.






                                  share|cite|improve this answer











                                  $endgroup$


















                                    0












                                    $begingroup$

                                    The answers provided already show many different ways this integral can be approached without Partial Fractions. Here is one more. Similar to the trigonometric substitution $u = sec(x)$ we can employ the hyperbolic substitution $u = tanh(x)$



                                    begin{align}
                                    int frac{1}{u^2 - 1}:du &= int frac{1}{tanh^2(x) - 1}cdot -operatorname{sech}^2(x):dx \
                                    &= int frac{1}{operatorname{sech}^2(x)}cdot -operatorname{sech}^2(x):dx \
                                    &= -x + C = -operatorname{arctanh(u)} + C
                                    end{align}



                                    Where $C$ is the constant of integration.






                                    share|cite|improve this answer











                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      The answers provided already show many different ways this integral can be approached without Partial Fractions. Here is one more. Similar to the trigonometric substitution $u = sec(x)$ we can employ the hyperbolic substitution $u = tanh(x)$



                                      begin{align}
                                      int frac{1}{u^2 - 1}:du &= int frac{1}{tanh^2(x) - 1}cdot -operatorname{sech}^2(x):dx \
                                      &= int frac{1}{operatorname{sech}^2(x)}cdot -operatorname{sech}^2(x):dx \
                                      &= -x + C = -operatorname{arctanh(u)} + C
                                      end{align}



                                      Where $C$ is the constant of integration.






                                      share|cite|improve this answer











                                      $endgroup$



                                      The answers provided already show many different ways this integral can be approached without Partial Fractions. Here is one more. Similar to the trigonometric substitution $u = sec(x)$ we can employ the hyperbolic substitution $u = tanh(x)$



                                      begin{align}
                                      int frac{1}{u^2 - 1}:du &= int frac{1}{tanh^2(x) - 1}cdot -operatorname{sech}^2(x):dx \
                                      &= int frac{1}{operatorname{sech}^2(x)}cdot -operatorname{sech}^2(x):dx \
                                      &= -x + C = -operatorname{arctanh(u)} + C
                                      end{align}



                                      Where $C$ is the constant of integration.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Jan 11 at 5:20

























                                      answered Jan 10 at 1:41







                                      user150203





































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