Prove

Let $x+2=A \frac{d}{d x}\left(4 x-x^{2}\right)+B$

$\Rightarrow x+2=A(4-2 x)+B$

Equating the coefficients of x and constant term on both sides, we obtain

$-2 A=1 \Rightarrow A=-\frac{1}{2}$

$4 A+B=2 \Rightarrow B=4$

$\Rightarrow(x+2)=-\frac{1}{2}(4-2 x)+4$

$\therefore \int \frac{x+2}{\sqrt{4 x-x^{2}}} d x$ $x=\int \frac{-\frac{1}{2}(4-2 x)+4}{\sqrt{4 x-x^{2}}} d x$

$=-\frac{1}{2} \int \frac{4-2 x}{\sqrt{4 x-x^{2}}} d x+4 \int \frac{1}{\sqrt{4 x-x^{2}}} d x$

Let $I_{1}=\int \frac{4-2 x}{\sqrt{4 x-x^{2}}} d x$ and $I_{2} \int \frac{1}{\sqrt{4 x-x^{2}}} d x$

$\therefore \int \frac{x+2}{\sqrt{4 x-x^{2}}} d x=-\frac{1}{2} I_{1}+4 I_{2}$   ...(1)

Then, $I_{1}=\int \frac{4-2 x}{\sqrt{4 x-x^{2}}} d x$

Let $4 x-x^{2}=t$

$\Rightarrow(4-2 x) d x=d t$

$\Rightarrow I_{1}=\int_{\sqrt{t}}^{d t}=2 \sqrt{t}=2 \sqrt{4 x-x^{2}}$    ...(2)

$I_{2}=\int \frac{1}{\sqrt{4 x-x^{2}}} d x$

$\Rightarrow 4 x-x^{2}=-\left(-4 x+x^{2}\right)$

$=\left(-4 x+x^{2}+4-4\right)$

$=4-(x-2)^{2}$

$=(2)^{2}-(x-2)^{2}$

$\therefore I_{2}=\int \frac{1}{\sqrt{(2)^{2}-(x-2)^{2}}} d x=\sin ^{-1}\left(\frac{x-2}{2}\right)$   ...(3)

Using equations (2) and (3) in (1), we obtain

$\int \frac{x+2}{\sqrt{4 x-x^{2}}} d x=-\frac{1}{2}\left(2 \sqrt{4 x-x^{2}}\right)+4 \sin ^{-1}\left(\frac{x-2}{2}\right)+\mathrm{C}$

$=-\sqrt{4 x-x^{2}}+4 \sin ^{-1}\left(\frac{x-2}{2}\right)+\mathrm{C}$