How can I calculate this indefinite integral: $\int\sqrt\frac{1}{x^3-1}~dx$

Question

Please help me to find this indefinite integration

$$\int\sqrt\frac{1}{x^3-1}~dx$$

I guess, this is an indefinite integral. Do you have a tip for me to solve this ? Thank you in advance!

Answer 1

Case $1$: $|x|\leq1$

Then $\int\sqrt{\dfrac{1}{x^3-1}}~dx$

$=\int\dfrac{1}{(x^3-1)^\frac{1}{2}}~dx$

$=\int\dfrac{1}{i(1-x^3)^\frac{1}{2}}~dx$

$=-i\int\sum\limits_{n=0}^\infty\dfrac{(2n)!x^{3n}}{4^n(n!)^2}~dx$

$=-\sum\limits_{n=0}^\infty\dfrac{i(2n)!x^{3n+1}}{4^n(n!)^2(3n+1)}+C$

Case $2$: $|x|\geq1$

Then $\int\sqrt{\dfrac{1}{x^3-1}}~dx$

$=\int\dfrac{1}{(x^3-1)^\frac{1}{2}}~dx$

$=\int\dfrac{1}{x^\frac{3}{2}(1-x^{-3})^\frac{1}{2}}~dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(2n)!x^{-3n-\frac{3}{2}}}{4^n(n!)^2}dx$

$=\sum\limits_{n=0}^\infty\dfrac{(2n)!x^{-3n-\frac{1}{2}}}{4^n(n!)^2\left(-3n-\dfrac{1}{2}\right)}+C$

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