Given integral
\(\begin{aligned}
& \int \frac{x^5 d x}{\left(x^2+x+1\right)\left(x^6+1\right)\left(x^4-x^3+x-1\right)} \\
& \because\left(x^2+x+1\right)\left(x^4-x^3+x-1\right)=(x-1)\left(x^3+1\right) \\
& \left(x^2+x+1\right) \\
& =\left(x^3-1\right)\left(x^3+1\right)=x^6-1 \\
& \therefore \text { Given integral }=\int \frac{x^5 d x}{\left(x^6+1\right)\left(x^6-1\right)} \\
& \text { put } x^6=t \Rightarrow 6 x^5 d x=d t
\end{aligned}\)
then given integral
\(\begin{aligned}
& =\frac{1}{6} \int \frac{d t}{(t+1)(t-1)}=\frac{1}{12} \log _e\left|\frac{t-1}{t+1}\right|+c \\
& =\frac{1}{12} \log _e\left|\frac{x^6-1}{x^6+1}\right|+c
\end{aligned}\)
Hence, option (2) is correct.