Integrate $\sin^{-1}\frac{2x}{1+x^2}$

You're right; but it depends on what the full question is; if one is asked to compute $$ \int_{0}^{1}\arcsin\frac{2x}{1+x^2}\,dx $$ then just the antiderivative $2\arctan x$ is sufficient. Not if one wants to compute an integral involving points outside the interval $[-1,1]$.

Here's a shorter way to get at your result.

Consider $$ f(x)=\arcsin\frac{2x}{1+x^2} $$ Then $$ f'(x)=\frac{1}{\sqrt{1-\dfrac{4x^2}{(1+x^2)^2}}}\frac{2(1+x^2)-4x^2}{(1+x^2)^2} =\frac{1+x^2}{|1-x^2|}\frac{2(1-x^2)}{(1+x^2)^2} $$ hence $$ f'(x)=\begin{cases} \dfrac{2}{1+x^2} & x\in(-1,1) \\[4px] -\dfrac{2}{1+x^2} & x\in(-\infty,-1)\cup(1,\infty) \end{cases} $$ Therefore, knowing that $f(0)=0$, $$ f(x)=\begin{cases} -\pi-2\arctan x & x<-1 \\[4px] 2\arctan x & -1\le x\le 1 \\[4px] \pi-2\arctan x & x>1 \end{cases} $$

In order to find an antiderivative we can consider $$ \int\arctan x\,dx=x\arctan x-\int\frac{x}{1+x^2}\,dx =x\arctan x-\frac{1}{2}\log(1+x^2) $$ Thus an antiderivative of $f$ has the form $$ F(x)=\begin{cases} c_- -\pi x-2x\arctan x+\log(1+x^2) &\qquad x<-1 \\[4px] 2x\arctan x-\log(1+x^2) &\qquad -1\le x\le 1 \\[4px] c_+ +\pi x-2x\arctan x+\log(1+x^2) &\qquad x>1 \end{cases} $$ and you just need to determine $c_-$ and $c_+$ to ensure continuity at $-1$ and $1$.

The other antiderivatives differ from $F$ by a constant.