The integral \(\displaystyle \int \frac{(x^8-x^2)\,dx}{\left(x^{12}+3x^6+1\right)\tan^{-1}\!\left(x^3+\frac{1}{x^3}\right)} \) is equal to:
\(\displaystyle \log_e\!\left(\Big|\tan^{-1}\!\left(x^3+\frac{1}{x^3}\right)\Big|\right)+C \)
\(\displaystyle \log_e\!\left(\Big|\tan^{-1}\!\left(x^3+\frac{1}{x^3}\right)\Big|\right)^{\displaystyle\frac{1}{3}}+C \)
\(\displaystyle \log_e\!\left(\Big|\tan^{-1}\!\left(x^3+\frac{1}{x^3}\right)\Big|\right)^{\displaystyle\frac{1}{2}}+C \)
\(\displaystyle \log_e\!\left(\Big|\tan^{-1}\!\left(x^3+\frac{1}{x^3}\right)\Big|\right)^{3}+C \)