I have tried to solve a problem here and I've got an answer but I just don't know if it is the right
Brenden Tran
Answered question
2022-06-22
I have tried to solve a problem here and I've got an answer but I just don't know if it is the right one. The problem is: For what conditions
represents a rational function. First I have calculated the integral and the result is:
where is a constant. From here I have taken the conditions
which is the same as
So, from my result, the condition above must be satisfied in order for that function to be a rational function. P.S != means not equal to Can anyone tell whether the solution is correct or not and if it is not then help me with it ?
Answer & Explanation
Dustin Durham
Beginner2022-06-23Added 31 answers
The only condition you want is that the coefficient in front of the logarithms are zero, the rest is considered to be a rational function (also, a polynomial is considered to be a rational function). In your case, this means that
(The term should be , but that does not change the answer.)
Mohamed Mooney
Beginner2022-06-24Added 5 answers
Since
it follows that the integral of over is rational iff, in the integrand function, the coefficient of , that is , and the coefficent of , that is are zero. So the condition should be simply