Now, if $u=x^{3}+6$, then $\int x^{2}\left(x^{3}+6\right)^{10} d x=\int u^{10}\left(\frac{1}{3} d u\right)=\frac{1}{3} \int u^{10} d u$.
This evaluates as
\[
\frac{1}{3} \int u^{10} d u=\frac{1}{3} \square+c=\square+C
\]

Step 1 ：Let \(u=x^{3}+6\), then \(\int x^{2}\left(x^{3}+6\right)^{10} d x=\int u^{10}\left(\frac{1}{3} d u\right)=\frac{1}{3} \int u^{10} d u\).

Step 2 ：The integral of \(u^{10}\) is \(\frac{1}{11}u^{11}\).

Step 3 ：Substitute \(u\) back in to get the final answer: \(u = x^{3} + 6\).

Step 4 ：The integral of \(x^{2}\left(x^{3}+6\right)^{10}\) with respect to \(x\) is \(\frac{x^{31}}{31} + \frac{15x^{28}}{7} + \frac{324x^{25}}{5} + \frac{12960x^{22}}{11} + \frac{272160x^{19}}{19} + 122472x^{16} + \frac{9797760x^{13}}{13} + 3359232x^{10} + \frac{75582720x^{7}}{7} + 25194240x^{4} + 60466176x + C\).

Step 5 ：\(\boxed{\text{Final Answer: The integral of } x^{2}\left(x^{3}+6\right)^{10} \text{ with respect to } x \text{ is } \frac{x^{31}}{31} + \frac{15x^{28}}{7} + \frac{324x^{25}}{5} + \frac{12960x^{22}}{11} + \frac{272160x^{19}}{19} + 122472x^{16} + \frac{9797760x^{13}}{13} + 3359232x^{10} + \frac{75582720x^{7}}{7} + 25194240x^{4} + 60466176x + C}\)