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

The solution is given in my reference as: $2x\tan^{-1}x-\log(1+x^2)+C$.
But, is it a complete solution ?
My Attempt
$$
\int 2\tan^{-1}x \, dx=\int \tan^{-1}x \cdot 2\,dx=\tan^{-1}x\int2\,dx-\int\frac{1}{1+x^2}\int2\,dx\cdot dx\\
=\tan^{-1}x \cdot2x-\int\frac{2x}{1+x^2}\,dx=2x\tan^{-1}x-\log(1+x^2)+C
$$
$$
2\tan^{-1}x=\begin{cases}\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|\leq{1}\\
\pi-\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|>{1}\text{ and }x>0\\
-\pi-\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|>{1}\text{ and }x<0
\end{cases}\\
\sin^{-1}\frac{2x}{1+x^2}=\begin{cases}2\tan^{-1}x\text{ if }|x|\leq{1}\\
\pi-2\tan^{-1}x\text{ if }|x|>{1}\text{ and }x>0\\
-\pi-2\tan^{-1}x\text{ if }|x|>{1}\text{ and }x<0
\end{cases}
$$
$$
\int\sin^{-1}\frac{2x}{1+x^2}\,dx=\begin{cases}\int2\tan^{-1}x\,dx&\text{ if } |x|\leq{1}\\\int\pi\, dx-\int2\tan^{-1}x\,dx&\text{ if }|x|>{1}&\text{ and } x>0\\-\int\pi \,dx-\int2\tan^{-1}x\,dx&\text{ if }|x|>{1}&\text{ and } x<0\end{cases}=\begin{cases}\color{red}{2x\tan^{-1}x-\log(1+x^2)+C\text{ if } |x|\leq{1}}\\\pi x-2x\tan^{-1}x+\log(1+x^2)+C&\text{ if }|x|>{1}&\text{ and } x>0\\-\pi x-2x\tan^{-1}x+\log(1+x^2)+C&\text{ if }|x|>{1}&\text{ and } x<0\end{cases}$$
So don't we have two more cases for our solution rather than that is given in my reference, right ?
 A: 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.
A: The "identity"
$$\arcsin\left(2x\over1+x^2\right)=2\arctan x$$
does not hold for all $x$.  You can see this by comparing the ranges of the two sides.  The arcsine function has $[-\pi/2,\pi/2]$ as its range, but twice the arctangent function has $(-\pi,\pi)$ as its range. In particular, if $x\gt1$, then $2\arctan x\gt\pi/2$, and likewise if $x\lt-1$ then $2\arctan x\lt-\pi/2$ (since $\arctan1=\pi/4$).
The correct identity is
$$\arcsin\left(2x\over1+x^2\right)=
\begin{cases}
\pi-2\arctan x\qquad\text{if }x\ge1\\
2\arctan x\qquad\quad\text{ }\text{ if }-1\le x\le1\\
-\pi-2\arctan x\quad\text{ if }x\le-1
\end{cases}$$
(Note, I've intentionally overlapped the intervals at $x=\pm1$ to stress the agreement of $\pi-2\arctan1=2\arctan1$ and $2\arctan(-1)=-\pi-2\arctan(-1)$.) The OP's observations regarding the integral follow.
A: According to the standard tangent half-angle substitution
\begin{align}
& x=\tan\frac\theta 2 \\[8pt]
& 2\arctan x = \theta \\[8pt]
& \frac {2\,dx}{1+x^2} = d\theta \\[8pt]
& dx = \sec^2\frac\theta 2 \,\,\frac{d\theta} 2 \\[8pt]
& \frac{2x}{1+x^2} = \sin\theta \\[8pt]
& \frac{1-x^2}{1+x^2} = \cos\theta
\end{align}
Thus we have $\theta = \arcsin\dfrac{2x}{1+x^2}.$
So
\begin{align}
& \int \arcsin\frac{2x}{1+x^2} \, dx = \int \theta \left( \sec^2\frac\theta 2 \,\, \frac{d\theta} 2 \right) = \int\theta \, dv = \theta v - \int v\,d\theta \\[10pt]
= {} & \theta\tan\frac\theta 2 - \int \tan\frac\theta 2 \, d\theta = \theta\tan\frac\theta 2 +2\log\left|\cos\frac\theta 2\right| + C \\[10pt]
= {} & \theta \tan \frac \theta 2 + \log\left|\frac 1 2 + \frac 1 2 \cos\theta \right| + C = \theta\tan\frac \theta 2 + \log\left| \frac 1 2 + \frac{1-x^2}{2(1+x^2)} \right| + C \\[10pt]
= {} & 2x\arctan x - \log(1+x^2) + C
\end{align}
A: We know that $\tan(\theta/2) = \frac{\sin(\theta)}{1+\cos(\theta)}$ and $\cos(\theta) = \sqrt{1-\sin^2(\theta)}$.
So, letting $x = \sin(\theta)$ $(\implies \theta = \arcsin(x))$ we have:
$\tan(\theta/2) = \frac{x}{1+\sqrt{1-x^2}} \implies \frac{\theta}{2} = \arctan\left(\frac{x}{1+\sqrt{1-x^2}}\right) \implies \arcsin(x) = 2\arctan\left(\frac{x}{1+\sqrt{1-x^2}}\right)$.
With that in hands, we just have to calculate $\arcsin\left(\frac{2x}{1+x^2}\right) = 2\arctan\left(\frac{\frac{2x}{1+x^2}}{1+\sqrt{1-\frac{4x^2}{1+2x^2+x^4}}}\right)$.
Now observe that $\frac{\frac{2x}{1+x^2}}{1+\sqrt{1-\frac{4x^2}{1+2x^2+x^4}}} = \frac{\frac{2x}{1+x^2}}{1+\sqrt{\frac{1-2x^2+x^4}{1+2x^2+x^4}}} = \frac{\frac{2x}{1+x^2}}{1+\frac{1-x^2}{1+x^2}} = \frac{\frac{2x}{1+x^2}}{\frac{2}{1+x²}} = x$. I thik this solve the equality you were struggling with.
