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 Explanation:

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 us substitute the term inside the logarithm:
$u={\tan }^{-1}(x^3+\frac{1}{x^3})$

Now, let us find the differential $du$ by applying the chain rule of differentiation:
$du=\frac{1}{1+{(x^3+\frac{1}{x^3})}^2}·\frac{d}{dx}(x^3+x^{-3})dx$

First, we simplify the denominator term $1+{(x^3+\frac{1}{x^3}}^2$:
$1+{(x^3+\frac{1}{x^3})}^2=1+x^6+2+\frac{1}{x^6}=x^6+3+\frac{1}{x^6}$
Taking the common denominator $x^6$, we get:
$1+{(x^3+\frac{1}{x^3})}^2=\frac{x^{12}+3x^6+1}{x^6}$

Next, we compute the derivative of the inner function:
$\frac{d}{dx}(x^3+x^{-3})=3x^2-3x^{-4}=3x^2-\frac{3}{x^4}=3\left(\frac{x^6-1}{x^4}\right)$

Now, substituting these back into the expression for $du$:
$du=\frac{x^6}{x^{12}+3x^6+1}·3\left(\frac{x^6-1}{x^4}\right)dx$
Simplifying the powers of $x$:
$du=\frac{3x^2(x^6-1)}{x^{12}+3x^6+1}dx$
$du=\frac{3(x^8-x^2)}{x^{12}+3x^6+1}dx$

Therefore, we can isolate the term present in our integral:
$\frac{(x^8-x^2)dx}{x^{12}+3x^6+1}=\frac{1}{3}du$

Substituting this back into the original integral:
$I=\int \frac{1}{3u}du=\frac{1}{3}\int \frac{1}{u}du$
Integrating $\frac{1}{u}$ gives:
$I=\frac{1}{3}\ln \left|u\right|+C$

Replacing $u$ with its original expression in terms of $x$:
$I=\frac{1}{3}\ln \left|{\tan }^{-1}(x^3+\frac{1}{x^3})\right|+C$