Evaluating indefinite integrals of the form $\int \frac{x^2 \,dx}{a x^5 + b}$

calculusindefinite-integralsintegrationpartial fractionsrational-functions

Evaluate the indefinite integral
$$\int \frac{x^2 \,dx}{a x^5 + b},$$
for real parameters $a, b \neq 0$.

No apparent substitutions simplify the expression (if the exponent of $x$ in the denominator were an integral multiple of $3$, the form of the integrand would suggest the substitution $u = x^3$, $du = 3 x^2 \,dx$, but the exponent is not). Applying integration by parts with $u = \frac{1}{a x^5 + b}$, $dv = x^2 dx$ is straightforward, but it produces an integrand with a much larger degree in the denominator and so appears only to make the situation worse. Applying integration by parts instead with $dw = \frac{x^k dx}{a x^5 + b}$ results in integrating $\frac{x^k dx}{a x^5 + b}$, which, except when $k \equiv 4 \pmod 5$ (which doesn't appear immediately fruitful), doesn't appear much easier than the given integral.

Best Answer

An explicit antiderivative is messy, but here's an outline for evaluating this integral by hand.

First, make a linear substitution $x = \alpha u$ for an appropriate constant $\alpha$, which transforms the integral $$\int \frac{x^2 \,dx}{a x^5 + b}$$ into some constant multiple of $$\int \frac{u^2 \,du}{1 - u^5} .$$ This is a rational expression, so in principle we can apply partial fractions and solve, but $1 - u^5$ factors over $\Bbb Q$ into $u - 1$ and a quartic polynomial irreducible over $\Bbb Q$. Thus, to factor the denominator into a product of linear and quadratic polynomials we need to resort to irrational coefficients.

Factoring a generic real quartic over $\Bbb R$ is unpleasant, but we can take advantage of the special form of the denominator: The roots of $1 - u^5$ are precisely the $5$th roots of unity, namely $1$ and the paired complex conjugates $e^{\pm 2 \pi i / 5}$ and $e^{\pm 4 \pi i / 5}$. Thus, one real quadratic factor of $1 - u^5$ is $$(u - e^{2 \pi i / 5}) (u - e^{-2 \pi i / 5}) = u^2 - 2 \cos \left(\frac{2 \pi}{5}\right) u + 1$$ and the other is $$(u - e^{4 \pi i / 5}) (u - e^{-4 \pi i / 5}) = u^2 + 2 \cos \left(\frac{\pi}{5}\right) u + 1 .$$ Optionally, we can rewrite these expressions using the facts that $2 \cos \frac{\pi}{5} = \phi$, where $\phi := \frac{1}{2}(1 + \sqrt{5})$ is the Golden Ratio, and $2 \cos \frac{2 \pi}{5} = \frac{1}{\phi}$.

Applying the Method of Partial Fractions thus gives a decomposition $$\frac{u^2}{1 - u^5} = \frac{A}{u - 1} + \frac{B u + C}{u^2 + \phi u + 1} + \frac{D u + E}{u^2 - \frac{1}{\phi} u + 1}$$ for some constants $A, B, C, D, E$, so $$\int \frac{u^2\, du}{1 - u^5} = A \int \frac{du}{u - 1} + \int \frac{(B u + C) du}{u^2 + \phi u + 1} + \int \frac{(D u + E) du}{u^2 - \frac{1}{\phi} u + 1} .$$

  • The integral $$\int \frac{du}{u - 1}$$ is elementary.
  • We can rewrite the integral of the second term as a linear combination of $$\int \frac{(2 u + \phi) du}{u^2 + \phi u + 1} \qquad \textrm{and} \qquad \int \frac{du}{u^2 + \phi u + 1} .$$ The left integral can be handled with the substitution $v = u^2 + \phi u + 1, dv = (2 u + \phi) du$, which gives $$\int \frac{(2 u + \phi) du}{u^2 + \phi u + 1} = \int \frac{dv}{v} = \log |v| + K = \log (u^2 + \phi u + 1) + K .$$ A linear substitution $w = \beta u + \gamma, dw = \beta \,du$ transforms the integral on the right into a multiple of $$\int \frac{dw}{w^2 + 1} = \arctan w + K' = \arctan (\beta u + \gamma) + K' .$$
  • The third integral can be handled much like the second integral.

With all of the integrals in $u$ now expressed in terms of elementary functions, all that remains is to undo the original substitution, that is, back-substitute $u = \frac{x}{\alpha}$ to produce an antiderivative in $x$.

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