Evaluate $\int\left(\sqrt{4-x^2}+x\right)dx$ I'm looking for simple way to solve

$$\int\left(\sqrt{4-x^2}+x\right) \, dx$$

I tried substitute $x=2\sin u$ and then
$$\cdots =\frac{x^2}{2}+\frac 1 2x\sqrt{4-x^2}+2\arcsin (x/2)+c$$
I'm looking for other solution please

Another way forward is to integrate by parts with $u=\sqrt{4-x^2}$ and $v=x$.  Then, we have
$$\begin{align}
\int \sqrt{4-x^2}\,dx&=x\sqrt{4-x^2}+\int \frac{x^2}{\sqrt{4-x^2}}\,dx\\\\
&=x\sqrt{4-x^2}+\int \frac{x^2-4+4}{\sqrt{4-x^2}}\,dx\\\\
&=x\sqrt{4-x^2}-\int \sqrt{4-x^2}\,dx+4\int \frac{1}{\sqrt{4-x^2}}\,dx\\\\
2 \int \sqrt{4-x^2}\,dx&=x\sqrt{4-x^2}+4\int \frac{1}{\sqrt{4-x^2}}\,dx\\\\
\int \sqrt{4-x^2}\,dx&=\frac12 x\sqrt{4-x^2}+2\int \frac{1}{\sqrt{4-x^2}}\,dx\\\\
&=\frac12 x\sqrt{4-x^2}+2\int \frac{1}{\sqrt{1-(x/2)^2}}\,d(x/2)\\\\
&=\frac12 x\sqrt{4-x^2}+2\arcsin(x/2)+C
\end{align}$$
