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

Question

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

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.

Answer 1

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}

Answer 2

$$\int\frac{1}{\left(1+x^2\right)\sqrt{1-\arctan^2(x)}}\space\text{d}x=$$ $$\int\frac{1}{\left(1+x^2\right)\sqrt{1-\left(\arctan(x)\right)^2}}\space\text{d}x=$$

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

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$$\int\frac{1}{\sqrt{1-u^2}}\space\text{d}u=$$ $$\arcsin\left(u\right)+\text{C}=$$ $$\arcsin\left(\arctan(x)\right)+\text{C}$$

Answer 3

Notice, $$\int \frac{dx}{(1+x^2)\sqrt{1-(\tan^{-1}(x))^2}}$$$$=\int \frac{\frac{dx}{1+x^2}}{\sqrt{1-(\tan^{-1}(x))^2}}$$$$=\int \frac{d(\tan^{-1}(x))}{\sqrt{1-(\tan^{-1}(x))^2}}$$$$=\sin^{-1}(\tan^{-1}(x))+C$$

Answer 4

Hint #1: Substitution: $$u=\arctan x$$

$$du=\frac{1}{1+x^2}dx$$

Hint #2: Derivative of arcsinx: $$\frac{d}{du}arcsin u=\frac{1}{\sqrt {1-u^2}}$$