The integral $\int <!--\; \int \; -->\frac{(x^8-<!--\; -\; -->x^2)dx}{(x^{12}+3x^6+1){\tan }^{-<!--\; -\; -->1}<!--\; \; -->(x^3+\frac{1}{x^3})}$ is equal to

The correct option is:
$$\frac{1}{3}\ln \left|{\tan }^{-1}(x^3+\frac{1}{x^3})\right|+C$$

Step-by-Step Derivation:

Let the given integral be:
$$I=\int \frac{(x^8-x^2)dx}{(x^{12}+3x^6+1){\tan }^{-1}(x^3+\frac{1}{x^3})}$$

To evaluate this integral, we can use the method of substitution. Let:
$$t={\tan }^{-1}(x^3+\frac{1}{x^3})$$

Now, differentiate both sides with respect to $x$ using the chain rule:
$$dt=\frac{1}{1+{(x^3+\frac{1}{x^3})}^2}\cdot \frac{d}{dx}(x^3+x^{-3})dx$$

We compute the derivative term and expand the square in the denominator:
$$dt=\frac{1}{1+(x^6+2+\frac{1}{x^6})}\cdot (3x^2-\frac{3}{x^4})dx$$

Simplify the terms:
$$dt=\frac{1}{x^6+3+\frac{1}{x^6}}\cdot 3\left(\frac{x^6-1}{x^4}\right)dx$$

Find a common denominator for the first fraction:
$$dt=\frac{x^6}{x^{12}+3x^6+1}\cdot \frac{3(x^6-1)}{x^4}dx$$

Combine the terms together:
$$dt=\frac{3x^2(x^6-1)}{x^{12}+3x^6+1}dx$$

Distribute $x^2$ in the numerator:
$$dt=\frac{3(x^8-x^2)}{x^{12}+3x^6+1}dx$$

This can be rewritten as:
$$\frac{(x^8-x^2)dx}{x^{12}+3x^6+1}=\frac{1}{3}dt$$

Substituting this result and the definition of $t$ back into the original integral $I$:
$$I=\int \frac{1}{3}\frac{dt}{t}$$

Integrating with respect to $t$:
$$I=\frac{1}{3}\ln \left|t\right|+C$$

Finally, substitute the value of $t$ back to get the solution in terms of $x$:
$$I=\frac{1}{3}\ln \left|{\tan }^{-1}(x^3+\frac{1}{x^3})\right|+C$$