Integral of constant divided by polynomial and another constant

Question

$$\int_{-\infty}^{-1}\frac{4}{\sqrt{x^6+2}}\,dx$$

What are the steps to integrate?

Answer 1

IF your upper limit would have been $0$ instead of $-1$, then we could easily have expressed this definite integral in terms of a $($regular$)$ beta function. As it stands, we can only do so by using an incomplete beta function instead. To do that, first rewrite the whole integral using the parity of its integrand, then factor $2$ outside of the radical sign. Then let $x^6=2~t^6$, and $u=\dfrac1{\sqrt{t^6+1}}$. The expression of the afore-mentioned special function will soon emerge.

Answer 2

$\int_{-\infty}^{-1}\dfrac{4}{\sqrt{x^6+2}}dx$

$=\int_\infty^1\dfrac{4}{\sqrt{(-x)^6+2}}d(-x)$

$=\int_1^\infty\dfrac{4}{\sqrt{x^6+2}}dx$

$=\int_1^\sqrt[6]2\dfrac{4}{\sqrt{x^6+2}}dx+\int_\sqrt[6]2^\infty\dfrac{4}{\sqrt{x^6+2}}dx$

$=\int_1^\sqrt[6]2\dfrac{2\sqrt2}{\sqrt{1+\dfrac{x^6}{2}}}dx+\int_\sqrt[6]2^\infty\dfrac{4}{x^3\sqrt{1+\dfrac{2}{x^6}}}dx$

$=\int_1^\sqrt[6]22\sqrt2\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{6n}}{4^n(n!)^22^n}dx+\int_\sqrt[6]2^\infty\dfrac{4}{x^3}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^n}{4^n(n!)^2x^{6n}}dx$

$=\int_1^\sqrt[6]2\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{6n}}{2^{3n-\frac{3}{2}}(n!)^2}dx+\int_\sqrt[6]2^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{-6n-3}}{2^{n-2}(n!)^2}dx$

$=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{6n+1}}{2^{3n-\frac{3}{2}}(n!)^2(6n+1)}\right]_1^\sqrt[6]2+\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{-6n-2}}{2^{n-2}(n!)^2(-6n-2)}\right]_\sqrt[6]2^\infty$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^{n+\frac{1}{6}}}{2^{3n-\frac{3}{2}}(n!)^2(6n+1)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{2^{3n-\frac{3}{2}}(n!)^2(6n+1)}+\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!2^{-n-\frac{1}{3}}}{2^{n-1}(n!)^2(3n+1)}$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{2^{2n-\frac{5}{3}}(n!)^2(6n+1)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{2^{3n-\frac{3}{2}}(n!)^2(6n+1)}+\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{2^{2n-\frac{2}{3}}(n!)^2(3n+1)}$