How to integrate $\int_{-1}^1 \tan^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right )\,dx$? Evaluate

$$\int_{-1}^1 \tan^{-1}\bigg (\dfrac{1}{\sqrt{1-x^2}}\bigg ) dx $$

Could somebody please help integrate this without using Differentiation under the Integral Sign?

First, integrate by parts to reduce the problem to calculating
$$\int \frac{x^2}{\sqrt{1-x^2}(2-x^2)}\,dx.$$
now split into two more manageable terms
$$\int \frac{2}{\sqrt{1-x^2}(2-x^2)}\,dx + \int \frac{x^2-2}{\sqrt{1-x^2}(2-x^2)}\,dx.$$
The left term is the only tricky one. Substitute $x= \sin u$ to get rid of the square root
$$\int \frac{2}{2-\sin^2 u}\,du = \int \frac{2}{2\cos^2 u+\sin^2 u}\,du = \int \frac{2\sec^2 u}{2+\tan^2 u}\,du$$
and finally substitute $v = \frac{\tan u}{\sqrt{2}}$.
