Let $u = x^3$. Then, $du = 3x^2dx$. We can see that $x^2dx$ is in the numerator, so this substitution will simplify the integral. We also need to rewrite $x^6$ as $(x^3)^2$ or $u^2$. So, the integral becomes
\[ \frac{1}{3} \int \frac{du}{u^2 + 3u + 2} \]
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