Evaluate $$ \int\left(2 x^{3}+1\right)^{7} x^{2} d x $$

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Calculate the integral $$ \int\left(2 x^{3}+1ight)^{7} x^{2} dx $$.
Evaluate the integral by following steps:
- step0: Evaluate using substitution:
  $$\int \left(2x^{3}+1ight)^{7}x^{2} dx$$
- step1: Use the substitution $$dx=\frac{1}{6x^{2}} dt$$ to transform the integral$$:$$
  $$\int \left(2x^{3}+1ight)^{7}x^{2}	imes \frac{1}{6x^{2}} dt$$
- step2: Simplify:
  $$\int \frac{\left(2x^{3}+1ight)^{7}}{6} dt$$
- step3: Use the substitution $$t=2x^{3}$$ to transform the integral$$:$$
  $$\int \frac{\left(t+1ight)^{7}}{6} dt$$
- step4: Use properties of integrals:
  $$\frac{1}{6}	imes \int \left(t+1ight)^{7} dt$$
- step5: Use the substitution $$dt=1 dv$$ to transform the integral$$:$$
  $$\frac{1}{6}	imes \int \left(t+1ight)^{7}	imes 1 dv$$
- step6: Simplify:
  $$\frac{1}{6}	imes \int \left(t+1ight)^{7} dv$$
- step7: Use the substitution $$v=t+1$$ to transform the integral$$:$$
  $$\frac{1}{6}	imes \int v^{7} dv$$
- step8: Evaluate the integral:
  $$\frac{1}{6}	imes \frac{v^{7+1}}{7+1}$$
- step9: Simplify:
  $$\frac{1}{6}	imes \frac{v^{8}}{8}$$
- step10: Multiply the terms:
  $$\frac{v^{8}}{6	imes 8}$$
- step11: Multiply the terms:
  $$\frac{v^{8}}{48}$$
- step12: Substitute back:
  $$\frac{\left(t+1ight)^{8}}{48}$$
- step13: Substitute back:
  $$\frac{\left(2x^{3}+1ight)^{8}}{48}$$
- step14: Add the constant of integral C:
  $$\frac{\left(2x^{3}+1ight)^{8}}{48} + C, C \in \mathbb{R}$$
The result of the integral $$ \int\left(2 x^{3}+1ight)^{7} x^{2} dx $$ is $$ \frac{(2x^{3}+1)^{8}}{48} + C $$, where $$ C $$ is an arbitrary constant.
$$ \frac{(2x^{3}+1)^{8}}{48} + C $$

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