How do I evaluate this improper integral $\int_{-1}^{1}\frac{dx}{(2-x)\sqrt{1-x^{2}}}$ Given integral is $$\int_{-1}^{1}\frac{dx}{(2-x)\sqrt{1-x^{2}}}.$$ I tried to split it up at $0$, but I donot know what to do ahead.
Thanks.

Let $\displaystyle I =\int_{-1}^{1}\frac{1}{(2-x)\sqrt{1-x^2}}dx....(1)$
Noe Let $x = -t\;,$ Then $dx = -dt$ and changing Limit, we get
$\displaystyle I = \int_{1}^{-1}\frac{1}{(2+t)\sqrt{1-t^2}}\times -dt = \int_{-1}^{1}\frac{1}{(2+x)\sqrt{1-x^2}}......(2)$
Now Add $(1)$ and $(2)\;,$ we get $\displaystyle 2I = \int_{-1}^{1}\left[\frac{1}{2-x}+\frac{1}{2+x}\right]\cdot \frac{1}{\sqrt{1-x^2}}dx = 4\int_{-1}^{1}\frac{1}{(4-x^2)\sqrt{1-x^2}}dx$
So $\displaystyle 2I = 8\int_{0}^{1}\frac{1}{(4-x^2)\sqrt{1-x^2}}dx$
Now Put $x=\sin \phi$ and $dx = \cos \phi\phi$ and Changing Limit, we get
$\displaystyle 2I = 8\int_{0}^{\frac{\pi}{2}}\frac{1}{4-\sin^2 \phi}d\phi. = \int_{0}^{\frac{\pi}{2}}\frac{\sec^2 \phi}{4(1+\tan^2 \phi)-1}d\phi = 8\int_{0}^{\frac{\pi}{2}}\frac{\sec^2 \phi}{3+4\tan^2 \phi}d\phi$
Now Put $\tan \phi = z\;,$ Then $\sec^2 \phi d\phi = dx$ and changing limit, we get
$\displaystyle 2I = 8\int_{0}^{\infty}\frac{1}{3+4z^2}dz = 2\int_{0}^{\infty}\frac{1}{z^2+\left(\frac{\sqrt{3}}{2}\right)^2}dz = \frac{4}{\sqrt{3}}\cdot \frac{\pi}{2}\Rightarrow I = \frac{\pi}{\sqrt{3}}$
