The integral  $\int \frac{(x^8-x^2)dx}{(x^{12}+3x^6+1)Ta{n}^{-1}(x^3+\frac{1}{x^3})}is$

$$\begin{gathered} Ta{n}^{-1}(x^3+\frac{1}{x^3})=t \\ \Rightarrow \frac{1}{1+{(x^3+\frac{1}{x^3})}^2}[3x^2-\frac{3}{x^4}]dx=dt \\ \Rightarrow \frac{x^6}{x^{12}+3x^6+1}\frac{3x^6-3}{x^4}dx=dt \\ I=\frac{1}{3}\int \frac{1}{t}dt=\frac{1}{3}\log \left|t\right|+c \\ =\log {\left|Ta{n}^{-1}(x^3+\frac{1}{x^3})\right|}^{\frac{1}{3}}+c \end{gathered}$$