8. $$ \int\left(2 x^{5}\right)\left(x^{2}-3\right)^{5} d x $$ $$ =\frac{1}{2 x} \int\left(2 x^{5}\right)\left(u^{5} d u\right. $$

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Let
$$
u=x^2-3.
$$
Then
$$
\frac{du}{dx}=2x\quad\Longrightarrow\quad du=2x\,dx.
$$

Since the integrand is
$$
2x^5 (x^2-3)^5\,dx,
$$
we rewrite $$x^5$$ as $$x^4 \cdot x$$ and use the substitution:
$$
x^5 dx = x^4\cdot x\,dx.
$$
Given that $$du=2x\,dx$$, we have
$$
x\,dx=\frac{du}{2}.
$$
Thus,
$$
2x^5 (x^2-3)^5\,dx=2x^4 (x^2-3)^5 \cdot \frac{du}{2}= x^4 (x^2-3)^5\,du.
$$

Now, express $$x^4$$ in terms of $$u$$. Notice that
$$
x^2=u+3,
$$
so
$$
x^4=(x^2)^2=(u+3)^2.
$$

The integral becomes
$$
\int x^4 (x^2-3)^5\,du=\int (u+3)^2 u^5\,du.
$$

Expand the product:
$$
(u+3)^2u^5=(u^2+6u+9)u^5= u^7+6u^6+9u^5.
$$

Thus, we have to integrate term by term:
$$
\int\left(u^7+6u^6+9u^5\right)du= \int u^7\,du+6\int u^6\,du+9\int u^5\,du.
$$

Perform the integrations:
$$
\int u^7\,du=\frac{u^8}{8},\quad 6\int u^6\,du=6\cdot\frac{u^7}{7}=\frac{6u^7}{7},\quad 9\int u^5\,du=9\cdot\frac{u^6}{6}=\frac{3u^6}{2}.
$$

Thus, the antiderivative is
$$
\frac{u^8}{8}+\frac{6u^7}{7}+\frac{3u^6}{2}+C.
$$

Finally, substitute back $$u=x^2-3$$:
$$
\boxed{\frac{(x^2-3)^8}{8}+\frac{6(x^2-3)^7}{7}+\frac{3(x^2-3)^6}{2}+C.}
$$
The integral simplifies to: $$ \frac{(x^2-3)^8}{8} + \frac{6(x^2-3)^7}{7} + \frac{3(x^2-3)^6}{2} + C $$

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