Integral of 1/(x^3 + bx - c) dx

[yuiop]

What is the integral of 1/(x^3 + bx - c) with respect to x?

This is part of a larger problem I am working on, but finding the integral has proved a sticking point. I have simplified it as much as possible.

P.S. The larger problem is finding J(r) in this PF post https://www.physicsforums.com/showpost.php?p=4397770&postcount=5

 

[HallsofIvy]

It will not be possible to get a simple form for that integral. To integrate "1 over a polynomial" you have to factor the polynomial. Here, you have a cubic so, theoretically, in terms of real numbers, either of two things can happen: you can factor it into 3 linear factors or you can factor it into one linear factor and one "irreducible" quadratic. If there are three linear factors, you can use "partial fractions" to write the integrand as the sum of three fractions with linear denominator so the integral is a combination of three logarithms. If there is on linear factor and one irreducible quadratic you can use "partial fractions" to write the integrand as the sum of one fraction with a linear denominator and one fraction with a linear numerator and quadratic denominator. The first integral is, of course, a logrithm. Since the quadratic denominator in the second fraction is irreducible, you can, by "completing the square", write the denominator in the form [itex](x- a)^2+ b[/itex]. Making the change of variable, u= x- a, you can then write the fraction as "(cu+d)/(u^2+ b)= cu/(u^2+ d)+ b/(u^2+ d). The first of those can be integrated by letting v= u^2+ d and the second as an arctangent.

However, there are simply too many choices and too many possible "routes" depending on the coefficients to be able to write a general formula.

 

[yuiop]

Since the quadratic denominator in the second fraction is irreducible, you can, by "completing the square", write the denominator in the form [itex](x- a)^2+ b[/itex].

It seams that in this case, I have an irreducible cubic so perhaps I need to 'complete the cube' but if I try that I end up with additional factors of x and x^2 outside the cube.

Is there another approach to this problem? In the linked post the integral has a term m(x) which is of course a function of x as defined in that post. Is it possible to carry out the integral as if m(x) was simply a constant (m) and then deal with m(x) after the integration has been carried out? It seams easy enough to obtain the integral of m(x) dx.