A calculus problem by A Former Brilliant Member

Theorem: $\int [g(x)]^n[g'(x)dx]=\dfrac{[g(x)]^{n+1}}{n+1}+c$ , where $n≠-1$

$\int x^2(5+2x^3)^8dx$

Observed that if $g(x)=5+2x^3$ then $g'(x)dx=6x^2dx$

Because

$\int x^2(5+2x^3)^8dx=\int (5+2x^3)^8(x^2dx)$

we need a factor of $6$ to go with $x^2$ to give $g'(x)dx$ . Therefore, we write

$\int (5+2x^3)^8(x^2dx)=\dfrac{1}{6} \int (5+2x^3)^8(6x^2dx)$

Applying the theorem above with $g(x)$ and $g'(x)dx$ given, we get

$\dfrac{1}{6} \int (5+2x^3)^8(6x^2dx)=\dfrac{1}{6} \dfrac{(5+2x^3)^9}{9}+C=$ $\color{#D61F06}\boxed{\dfrac{1}{54}(5+2x^3)^9+C}$