Ok, so, as above:

$u = 4x-x^2, du = (4-2x)dx$

$u(2) = 4(2) - (2)^2 = 4$

$u(3) = 4(3) - (3)^2 = 3$

So,

$\displaystyle \begin{align*}\int_2^3 \dfrac{2x-3}{\sqrt{4x-x^2}}dx & = -\int_{u(2)}^{u(3)}u^{-1/2}du + \int_2^3 \dfrac{1}{\sqrt{4x-x^2}}dx \\ & = -\int_4^3 u^{-1/2}du + \int_2^3 \dfrac{1}{\sqrt{4x-x^2}}dx \\ & = \left. -2\sqrt{u} \right]_4^3 + \int_2^3 \dfrac{1}{\sqrt{4x-x^2}}dx \\ & = (-2\sqrt{3}+4) + \int_2^3 \dfrac{1}{\sqrt{4x-x^2}}dx \end{align*}$

Now, to complete the square for the second integral, you have

$\displaystyle \begin{align*}4x-x^2 & = -(x^2-4x) \\ & = -\left(x^2-4x + \left(\frac{4}{2} \right)^2 - \left(\frac{4}{2} \right)^2\right) \\ & = -\left(x^2-4x+4-4\right) \\ & = -\left((x-2)^2-4\right) \\ & = 4-(x-2)^2\end{align*}$

So, you have:

$\displaystyle \begin{align*}(-2\sqrt{3}+4) + \int_2^3 \dfrac{1}{\sqrt{4x-x^2}}dx & = (-2\sqrt{3}+4) + \int_2^3 \dfrac{1}{\sqrt{4-(x-2)^2}}dx \\ & = (-2\sqrt{3}+4) + \int_2^3 \dfrac{1}{\sqrt{4\left(1-\left(\dfrac{x-2}{2}\right)^2 \right) }}dx \\ & = (-2\sqrt{3}+4) + \int_2^3 \dfrac{1}{\sqrt{1-\left(\dfrac{x-2}{2} \right)^2}}\left(\dfrac{1}{2} \right)dx\end{align*}$

The final equality comes from taking the square root of 4 in the denominator and using commutativity and associativity to move it next to the $dx$.

So, let $u = \dfrac{x-2}{2}$. Then $du = \dfrac{1}{2}dx$. This means $u(2) = 0$ and $u(3) = \dfrac{1}{2}$. So, you have:

$\displaystyle \begin{align*}(-2\sqrt{3} + 4) + \int_2^3 \dfrac{1}{\sqrt{1-\left( \dfrac{x-2}{2} \right)^2}} \left(\dfrac{1}{2} \right)dx & = (-2\sqrt{3}+4) + \int_{u(2)}^{u(3)} \dfrac{1}{\sqrt{1-u^2}}du \\ & = (-2\sqrt{3}+4) + \left. \arcsin(u) \right]_0^{1/2} \\ & = (-2\sqrt{3}+4) + \left(\dfrac{\pi}{6}-0\right) \\ & = 4+\dfrac{\pi}{6}-2\sqrt{3}\end{align*}$

Here it is through WolframAlpha (

link). Wolframalpha does the same problem with trigonometric substitution which makes it much easier to solve.