Solving $\int {dx\over(1+x^2)\sqrt{1-(\arctan x)^2}}$ 
$$\int {dx\over (1+x^2)\sqrt{1-(\arctan x)^2}}$$

I found a similar answer but did not manage to use it in this case
I can see the function $\arcsin x'$ and $\arctan x'$ in the expression.
 A: Set $\arctan(x) = t$. We then have $\tan(t) = x \implies dx = \sec^2(t)dt$. Hence, the integral becomes
\begin{align}
\int \dfrac{dx}{(1+x^2)\sqrt{(1-(\arctan(x))^2)}} & = \int \dfrac{dt}{\sqrt{1-t^2}} = \arcsin(t) + \text{const} = \arcsin\left(\arctan(x)\right) + \text{const}
\end{align}
A: $$\int\frac{1}{\left(1+x^2\right)\sqrt{1-\arctan^2(x)}}\space\text{d}x=$$
$$\int\frac{1}{\left(1+x^2\right)\sqrt{1-\left(\arctan(x)\right)^2}}\space\text{d}x=$$

Substitute $u=\arctan(x)$ and $\text{d}u=\frac{1}{x^2+1}\space\text{d}x$:

$$\int\frac{1}{\sqrt{1-u^2}}\space\text{d}u=$$
$$\arcsin\left(u\right)+\text{C}=$$
$$\arcsin\left(\arctan(x)\right)+\text{C}$$
A: Notice, $$\int \frac{dx}{(1+x^2)\sqrt{1-(\tan^{-1}(x))^2}}$$$$=\int \frac{\frac{dx}{1+x^2}}{\sqrt{1-(\tan^{-1}(x))^2}}$$$$=\int \frac{d(\tan^{-1}(x))}{\sqrt{1-(\tan^{-1}(x))^2}}$$$$=\sin^{-1}(\tan^{-1}(x))+C$$
A: 
Hint #1: Substitution: $$u=\arctan x$$
$$du=\frac{1}{1+x^2}dx$$
Hint #2: Derivative of arcsinx: $$\frac{d}{du}arcsin u=\frac{1}{\sqrt {1-u^2}}$$

