Integral $\int \sqrt{\frac{x}{2-x}}dx$

$$\int \sqrt{\frac{x}{2-x}}dx$$

Set $t=\frac {x} {2-x}$ and $dt=\left(\frac{x}{(2-x)^2}+\frac{1}{2-x}\right)dx$

$$=2\int\frac{\sqrt t}{(t+1)^2}dt$$

Set $\nu=\sqrt t$ and $d\nu=\frac{dt}{2\sqrt t}$

$$=4\int\frac{\nu^2}{(\nu^2+1)^2}d\nu\overset{\text{ partial fractions}}{=}4\int\frac{d\nu}{\nu^2+1}-4\int\frac{d\nu}{(\nu^1+1)^2+\mathcal C}$$

$$=4\arctan \nu-4\int\frac{d\nu}{(\nu^2+1)^2}$$

Set $\nu=\tan p$ and $d\nu=\sec^2 p dp.$ Then $(\nu^2+1)^2=(\tan^2 p+1)^2=\sec^4 p$ and $p=\arctan \nu$

$$=4\arctan \nu-4\int \cos^2 p dp$$

$$=4\arctan \nu-2\int \cos(2p)dp-2\int 1dp$$

$$=4\arctan \nu-\sin(2p)-2p+\mathcal C$$

Set back $p$ and $\nu$:

$$=\color{red}{\sqrt{-\frac{x}{x-2}}(x-2)+2\arctan\left(\sqrt{-\frac{x}{x-2}}\right)+\mathcal C}$$

Let me try do derive that antiderivative. You computed:

$$f(x)=\underbrace{-2\arcsin\sqrt{\frac{2-x}{2}}}_{f_1(x)}\underbrace{-\sqrt{2x-x^2}}_{f_2(x)}.$$

The easiest term is clearly $f_2$:

$$f_2'(x)=-\frac{1}{2\sqrt{2x-x^2}}\frac{d}{dx}(2x-x^2)=\frac{x-1}{\sqrt{2x-x^2}}.$$

Now the messier term. Recall that $\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}}$. So:

\begin{align*} f_1'(x)={}&-2\frac{1}{\sqrt{1-\left(\sqrt{\frac{2-x}{2}}\right)^2}}\frac{d}{dx}\sqrt{\frac{2-x}{2}}=-2\frac{1}{\sqrt{1-\frac{2-x}{2}}}\cdot\frac{1}{\sqrt2}\frac{d}{dx}\sqrt{2-x}={} \\ {}={}&-2\sqrt{\frac2x}\cdot\frac{1}{\sqrt2}\cdot\frac{1}{2\sqrt{2-x}}\cdot(-1)=\frac{2}{\sqrt x}\frac{1}{2\sqrt{2-x}}=\frac{1}{\sqrt{2x-x^2}}. \end{align*}

So:

$$f'(x)=f_1'(x)+f_2'(x)=\frac{x}{\sqrt{2x-x^2}}=\frac{x}{\sqrt x}\frac{1}{\sqrt{2-x}}=\frac{\sqrt x}{\sqrt{2-x}},$$

which is your integrand. So you were correct after all! Or at least got the correct result, but no matter how I try, I cannot find an error in your calculations.

As for the book's solution, take your $f$, and compose it with $g(x)=2-x$. You get the book's solution, right? Except for a sign. But then $g'(x)=-1$, so the book's solution is also correct: just a different change of variables, probably, though I cannot really guess which.