Вычислить: \int_{2}^{3} x(3-x)^7 dx

Let $$u = 3-x$$, then $$x = 3-u$$ and $$du = -dx$$. When $$x=2$$, $$u=1$$. When $$x=3$$, $$u=0$$. The integral becomes:
\int_{1}^{0} (3-u)u^7 (-du) = \int_{0}^{1} (3u^7 - u^8)du = \left[ \frac{3u^8}{8} - \frac{u^9}{9} \right]_{0}^{1} = \left( \frac{3(1)^8}{8} - \frac{(1)^9}{9} \right) - (0) = \frac{3}{8} - \frac{1}{9} = \frac{27-8}{72} = \frac{19}{72}