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Recently I have encountered the integral of a rational function, which arose upon evaluating the ordinary integral of an irrational function, namely $\int \sqrt[n]{\frac{ax + b}{cx + d}}dx$:

$$I =\int \frac{dx}{x^n -a^n}.$$

It can be observable that by using the formula

$$x^n - a^n = (x -a)(x^{n-1} +ax^{n-2} + a^2 x^{n-3}+...+a^{n-2}x + a^{n-1}),$$

the integrand can be represented as a sum of several fractions according to the unknown coefficients method; in doing so, one arrives at the sum of two integrals, and the latter of which is more tougher for the reason that a polynomial of the $(n-1)-$th power is in the denominator of the acquired integrand:

$$I= \int \frac{Adx}{x-a} +\int \frac{P_{n-2} (x)dx}{x^{n-1} +ax^{n-2} + a^2 x^{n-3}+...+a^{n-2}x + a^{n-1}}$$

Because the integral of the nominator, $P_{n-2} (x)$, equals to a new polynomial whose the power is raised by 1, that is, $Q_{n - 1} (x)$. Therefore, the differential of the polynomial is the one resembling $P_{n - 2} (x)$. Here the idea on how to evaluate the integral occurs:

$$d(x^{n-1} +ax^{n-2} + a^2 x^{n-3}+...+a^{n-2}x + a^{n-1}) = P_{n-2} (x)+\lambda,$$

where the last term is another polynomial whose power is less than that of $P_{n-2} (x)$. Despite what has been done, the problem is not solved. Why? Because one has produced another perplexing integral:

$$J = \int \frac{\lambda_{p} (x)dx}{x^{n-1} +ax^{n-2} + a^2 x^{n-3}+...+a^{n-2}x + a^{n-1}},$$

where $0 < p < n-2$. To discover the path to complete the solution, it is more reasonable to consider one representative of the integrals type, for example,

$$ J' =\int \frac{x^2 +x + 1}{x^4 + x^2 + x - 10}dx $$

Provided that the denominator cannot be factorised, one cannot find the antiderivative easily. Otherwise, the fraction is again decomposed; yet, it remains unclear as to how to do the integrals of the type. Should they not be able to be expressed in terms of elementary functions combinations, then the question of how to prove this appears.

sergei ivanov
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    With complex numbers, you get partial fractions with denominators of degree $1$, hence an explicit expression for the integral (as a linear combination of logs), convertible to a purely real one (if desired). Why complicate?.. – metamorphy Oct 16 '19 at 20:14

1 Answers1

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You can decompose any polynomial with real coefficients, with degree greater than 2 into a product of polynomials of degree 1 and 2, all with real coefficients. For example: $$x^2+x^2+x-10=(x^2+ax+b)(x^2+cx+d)$$ Identifying coefficients, I've got $a=-c\approx 2.13$, $b\approx 2.95$, $d3.38$. To get $a$ I've used numerical methods. For the rest it's simple arithmetic. Once should check if any of these quadratics can be further decomposed into linear functions (meaning checking if the roots are real).

Andrei
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  • This is obvious. The real question is what to do when having no real roots. – sergei ivanov Oct 16 '19 at 20:01
  • There are simple ways to calculate $\int\frac{x+a}{x^2+bx+c}dx$ and $\int\frac{1}{x^2+bx+c}dx$. They rely on completing the square and change variables to get something like $\int\frac{2y}{y^2+1} dy=\int\frac{dz}z=\ln z$ or $\int\frac{1}{y^2+1} dy=\arctan y$ – Andrei Oct 16 '19 at 21:23
  • Were the question to be devoted to these integrals, I would have the problem solved. The matter is that of integrals $P_{2} (x)$/$Q_{4} (x)$. – sergei ivanov Oct 16 '19 at 22:30
  • The reason why your A has been accepted is due to me having realized that the method of fraction decomposition works even if all roots are complex, meaning that the polynomial is still factorizable. Hence, I can split the fraction as long as the degree of the numerator is less than that of the denominator. However, that does not help you evaluat the general integral $\int \frac{1}{x^n - a^n}dx$. Why? Because there is a lot of work but the idea is the same. – sergei ivanov Jan 24 '21 at 01:28