Is this approach correct? Can the result be simplified?
I got to:
$$\frac{3}{2}\int\sqrt{1-(\frac{2x-1}{3})^2}\mathrm{dx},$$
substituted $\frac{2x-1}{3}=\cos(y)$, and got a result:
$$-\frac{9}{8}\arccos(\frac{2x-1}{3})+\frac{9}{16}\sin(2\arccos(\frac{2x-1}{3}))+C$$
On
$$ \begin{aligned} I &= \int\sqrt{2+x-x^2}dx \\ &\stackrel{\text{integration by parts}}{=} x\sqrt{2+x-x^2} - \int xd\sqrt{2+x-x^2} \\ &= x\sqrt{2+x-x^2} - \int x\frac{1-2x}{2\sqrt{2+x-x^2}}dx \\ &= x\sqrt{2+x-x^2} - \int \frac{x-2x^2}{2\sqrt{2+x-x^2}}dx \\ &= x\sqrt{2+x-x^2} - \frac{1}{2}\int \frac{4+2x-2x^2-(4+x)}{\sqrt{2+x-x^2}}dx \\ &= x\sqrt{2+x-x^2} - \int \sqrt{2+x-x^2}dx + \frac{1}{2} \int \frac{4+x}{\sqrt{2+x-x^2}} \\ &= x\sqrt{2+x-x^2} -I + \frac{1}{2} \int \frac{4+x}{\sqrt{2+x-x^2}}dx \Rightarrow \end{aligned} $$
$$ \begin{aligned} 2I &= x\sqrt{2+x-x^2} + \frac{1}{2} \int \frac{4+x}{\sqrt{2+x-x^2}}dx = \\ &= x\sqrt{2+x-x^2} - \frac{1}{2} \int \frac{1-2x}{2\sqrt{2+x-x^2}}dx + \frac{9}{4} \int \frac{1}{\sqrt{2+x-x^2}}dx\\ &= x\sqrt{2+x-x^2} - \frac{1}{2} \int d\sqrt{2+x-x^2} + \frac{9}{4} \int \frac{1}{\sqrt{\left(\frac{3}{2}\right)^2-\left(x-\frac{1}{2}\right)^2}}dx\\ &=x\sqrt{2+x-x^2} - \frac{1}{2}\sqrt{2+x-x^2} + \frac{9}{4}\arcsin\left(\frac{\left(x-\frac{1}{2}\right)}{\frac{3}{2}}\right) +C \Rightarrow \\ \\ I &= \frac{1}{2}x\sqrt{2+x-x^2} - \frac{1}{4}\sqrt{2+x-x^2} + \frac{9}{8}\arcsin\left(\frac{2x-1}{3}\right) +C \end{aligned} $$
Any two antiderivatives should only differ by a constant, by the FTC. The double-angle formulas are sometimes helpful to simplify things once you've found an antiderivative. In particular, $$ \sin(2\theta) = 2\sin(\theta)\cos(\theta) $$So something like $\sin(2\arccos(\theta))=2\sin(\arccos(\theta))\cos(\arccos(\theta))$. The second term reduces to $\theta$ and the first term simplifes to $\sqrt{1-\theta^2}$ by drawing a triangle.