Integrate the function: $\frac{x+2}{\sqrt{4x-x^2}}$

(N/A) Let $x+2 = A \frac{d}{dx}(4x-x^2) + B$.
$\Rightarrow x+2 = A(4-2x) + B$.
Equating the coefficients of $x$ and the constant term on both sides,we obtain:
$-2A = 1 \Rightarrow A = -\frac{1}{2}$.
$4A + B = 2 \Rightarrow B = 4$.
Thus,$x+2 = -\frac{1}{2}(4-2x) + 4$.
Therefore,$\int \frac{x+2}{\sqrt{4x-x^2}} dx = \int \frac{-\frac{1}{2}(4-2x) + 4}{\sqrt{4x-x^2}} dx$.
$= -\frac{1}{2} \int \frac{4-2x}{\sqrt{4x-x^2}} dx + 4 \int \frac{1}{\sqrt{4x-x^2}} dx$.
Let $I_1 = \int \frac{4-2x}{\sqrt{4x-x^2}} dx$ and $I_2 = \int \frac{1}{\sqrt{4x-x^2}} dx$.
So,the integral is $-\frac{1}{2}I_1 + 4I_2$ $(1)$.
For $I_1$,let $4x-x^2 = t$,then $(4-2x)dx = dt$.
$I_1 = \int \frac{dt}{\sqrt{t}} = 2\sqrt{t} = 2\sqrt{4x-x^2}$ $(2)$.
For $I_2$,$4x-x^2 = -(x^2-4x) = -(x^2-4x+4-4) = 4-(x-2)^2 = (2)^2-(x-2)^2$.
$I_2 = \int \frac{dx}{\sqrt{(2)^2-(x-2)^2}} = \sin^{-1}\left(\frac{x-2}{2}\right)$ $(3)$.
Substituting $(2)$ and $(3)$ into $(1)$:
$= -\frac{1}{2}(2\sqrt{4x-x^2}) + 4\sin^{-1}\left(\frac{x-2}{2}\right) + C$.
$= -\sqrt{4x-x^2} + 4\sin^{-1}\left(\frac{x-2}{2}\right) + C$,where $C$ is the constant of integration.