9) $$\int_{-1}^{2}(x^2-1)^3xdx$$.

Let $$u = x^2 - 1$$. Then $$du = 2x , dx$$, so $$x , dx = \frac{1}{2} , du$$.

When $$x = -1$$, $$u = (-1)^2 - 1 = 1 - 1 = 0$$.

When $$x = 2$$, $$u = (2)^2 - 1 = 4 - 1 = 3$$.

The integral becomes $$ int_0^3 u^3 \frac{1}{2} , du = \frac{1}{2} int_0^3 u^3 , du $$

$$ \frac{1}{2} int_0^3 u^3 , du = \frac{1}{2} \left[ \frac{u^4}{4} \right]_0^3 = \frac{1}{2} \left( \frac{3^4}{4} - \frac{0^4}{4} \right) = \frac{1}{2} \cdot \frac{81}{4} = \frac{81}{8} $$

So, $$\int_{-1}^{2}(x^2-1)^3xdx = \frac{81}{8}$$.

Answer: $$\frac{81}{8}$$