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
$$
\int 2\tan^{-1}x \, dx=\int \tan^{-1}x \cdot 2\,dx=\tan^{-1}x\int2\,dx-\int\frac{1}{1+x^2}\int2\,dx\cdot dx\\
=\tan^{-1}x \cdot2x-\int\frac{2x}{1+x^2}\,dx=2x\tan^{-1}x-\log(1+x^2)+C
$$
$$
2\tan^{-1}x=\begin{cases}\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|\leq{1}\\
\pi-\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|&gt;{1}\text{ and }x&gt;0\\
-\pi-\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|&gt;{1}\text{ and }x&lt;0
\end{cases}\\
\sin^{-1}\frac{2x}{1+x^2}=\begin{cases}2\tan^{-1}x\text{ if }|x|\leq{1}\\
\pi-2\tan^{-1}x\text{ if }|x|&gt;{1}\text{ and }x&gt;0\\
-\pi-2\tan^{-1}x\text{ if }|x|&gt;{1}\text{ and }x&lt;0
\end{cases}
$$
$$
\int\sin^{-1}\frac{2x}{1+x^2}\,dx=\begin{cases}\int2\tan^{-1}x\,dx&amp;\text{ if } |x|\leq{1}\\\int\pi\, dx-\int2\tan^{-1}x\,dx&amp;\text{ if }|x|&gt;{1}&amp;\text{ and } x&gt;0\\-\int\pi \,dx-\int2\tan^{-1}x\,dx&amp;\text{ if }|x|&gt;{1}&amp;\text{ and } x&lt;0\end{cases}=\begin{cases}\color{red}{2x\tan^{-1}x-\log(1+x^2)+C\text{ if } |x|\leq{1}}\\\pi x-2x\tan^{-1}x+\log(1+x^2)+C&amp;\text{ if }|x|&gt;{1}&amp;\text{ and } x&gt;0\\-\pi x-2x\tan^{-1}x+\log(1+x^2)+C&amp;\text{ if }|x|&gt;{1}&amp;\text{ and } x&lt;0\end{cases}$$

So don't we have two more cases for our solution rather than that is given in my reference, right ?
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
$$
f'(x)=\begin{cases}
\dfrac{2}{1+x^2} &amp; x\in(-1,1) \\[4px]
-\dfrac{2}{1+x^2} &amp; x\in(-\infty,-1)\cup(1,\infty)
\end{cases}
$$
Therefore, knowing that $f(0)=0$,
$$
f(x)=\begin{cases}
-\pi-2\arctan x &amp; x&lt;-1 \\[4px]
2\arctan x &amp; -1\le x\le 1 \\[4px]
\pi-2\arctan x &amp; x&gt;1
\end{cases}
$$

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
$$
F(x)=\begin{cases}
c_- -\pi x-2x\arctan x+\log(1+x^2) &amp;\qquad x&lt;-1 \\[4px]
2x\arctan x-\log(1+x^2) &amp;\qquad -1\le x\le 1 \\[4px]
c_+ +\pi x-2x\arctan x+\log(1+x^2) &amp;\qquad x&gt;1
\end{cases}
$$
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.
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.