The integral $$\int \frac{x^8 - x^2}{(x^{12} + 3x^6 + 1)\tan^{-1}\left(x^3 + \frac{1}{x^3}\right)} dx$$ is equal to :

Let $$u = \tan^{-1}\left(x^3 + \frac{1}{x^3}\right)$$.

$$du = \frac{1}{1 + (x^3 + x^{-3})^2} \cdot (3x^2 - 3x^{-4})dx = \frac{3(x^2 - x^{-4})}{1 + (x^3+x^{-3})^2}dx$$

Note: $$1 + (x^3 + x^{-3})^2 = 1 + x^6 + 2 + x^{-6} = x^6 + 3 + x^{-6}$$

$$= \frac{x^{12} + 3x^6 + 1}{x^6}$$

$$du = \frac{3(x^2 - x^{-4}) \cdot x^6}{x^{12} + 3x^6 + 1}dx = \frac{3(x^8 - x^2)}{x^{12} + 3x^6 + 1}dx$$

So $$\frac{x^8 - x^2}{x^{12} + 3x^6 + 1}dx = \frac{du}{3}$$.

The integral becomes:

$$\int \frac{1}{u} \cdot \frac{du}{3} = \frac{1}{3}\ln|u| + C = \frac{1}{3}\ln\left|\tan^{-1}\left(x^3+\frac{1}{x^3}\right)\right| + C$$

$$= \ln\left[\tan^{-1}\left(x^3+\frac{1}{x^3}\right)\right]^{1/3} + C$$

The answer corresponds to Option (1).