$\int_0^2 x(2-x)^n dx$ is equal to

Let $u = 2-x$

Then $x = 2-u$

$du = -dx$, so $dx = -du$

When $x = 0$, $u = 2$

When $x = 2$, $u = 0$

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$\int_0^2 x(2-x)^n dx = \int_2^0 (2-u) \cdot u^n \cdot (-du)$

$= \int_0^2 (2-u) u^n \, du$

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$\int_0^2 (2-u) u^n \, du = \int_0^2 (2u^n - u^{n+1}) \, du$

$= \int_0^2 2u^n \, du - \int_0^2 u^{n+1} \, du$

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$= 2 \cdot \left[\frac{u^{n+1}}{n+1}\right]_0^2 - \left[\frac{u^{n+2}}{n+2}\right]_0^2$

$= 2 \cdot \frac{2^{n+1}}{n+1} - \frac{2^{n+2}}{n+2}$

$= \frac{2^{n+2}}{n+1} - \frac{2^{n+2}}{n+2}$

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$= 2^{n+2} \left(\frac{1}{n+1} - \frac{1}{n+2}\right)$

$= 2^{n+2} \left(\frac{(n+2) - (n+1)}{(n+1)(n+2)}\right)$

$= 2^{n+2} \left(\frac{1}{(n+1)(n+2)}\right)$

$= \frac{2^{n+2}}{(n+1)(n+2)}$