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

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 \(u=x^2-3\): \[ \boxed{\frac{(x^2-3)^8}{8}+\frac{6(x^2-3)^7}{7}+\frac{3(x^2-3)^6}{2}+C.} \]