Let $u = 1 + x^2$. Then, $du = 2x dx$. Notice that we have $x^6 dx$ in the integrand, so we can rewrite it in terms of $u$ by solving for $x^6 dx$:
$x^6 dx = \frac{1}{2} (u - 1)^3 du$
Now, our integral becomes:
$$
\int \frac{x^6}{1+x^2} dx = \frac{1}{2} \int
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