$$I=\int \frac{x^8-x^2}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)} d x$$

$$\begin{aligned} & \text { Let } \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)=t \\ & \Rightarrow \frac{1}{1+\left(x^3+\frac{1}{x^3}\right)^2} \cdot\left(3 x^2-\frac{3}{x^4}\right) d x=d t \\ & \Rightarrow \frac{x^6}{x^{12}+3 x^6+1} \cdot \frac{3 x^6-3}{x^4} d x=d t \\ & I=\frac{1}{3} \int \frac{d t}{t}=\frac{1}{3} \ln |t|+C \\ & I=\frac{1}{3} \ln \left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|+C \\ & I=\ln \left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|^{1 / 3}+C \end{aligned}$$

Hence option (1) is correct.