Indefinite integral question with cube roots Evaluate the following integral:
$$\int\sqrt[3]{x^3+\frac 1 {x^3}} dx$$
For $x^3=u$ and $1+\frac 1 {u^2}=v$, getting $$-\frac 1 6 \int(v-1)^{-\frac 4 3}v^{\frac 1 3} dv$$, what is the next step?

I would chose a different substitution:
$$I =\int \big(x^3 +\frac{1}{x^3} \big)^{\frac{1}{3}}dx$$
$$=\int \frac{(x^6 +1)^{\frac{1}{3}}}{x} dx $$
Let: $ u=(x^6+1)^{\frac{1}{3}} $, then:
$x^6=u^3-1 \text{ , and } dx= \frac{(x^6 +1)^{\frac{2}{3}}}{2x^5} du$ .
$$I=\frac{1}{2} \int \frac{u^3}{u^3-1}du$$
$$=\frac{1}{2} \int \frac{1}{u^3-1}du \text{ }+ \text{ } \frac{1}{2}u$$
The first integral can easily be solved now with partial fraction decomposition; and using the derivative of the arctan function.
