How do I solve $\int_{-1}^1 \frac{1}{(x-2)(x+2)\sqrt{1-x^2}} dx$? I'm hopelessly stuck with the integral
$$\int_{-1}^1 \frac{1}{(x-2)(x+2)\sqrt{1-x^2}} dx$$
I'm guessing the way to go is complex contour integration, but I have no idea.
 A: Let $x=\sin t$
$$\int_{-1}^1 \frac{1}{(x-2)(x+2)\sqrt{1-x^2}} dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1}{\sin^2t-4}dt$$
and then $\sin t=\dfrac{2\tan\frac{t}{2}}{1+\tan\frac{t}{2}}$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
&\int_{-1}^{1}{\dd x \over \pars{x - 2}\pars{x + 2}\root{1 - x^{2}}} =
2\int_{0}^{1}{\dd x \over \pars{x^{2} - 4}\root{1 - x^{2}}}
\\[5mm] \stackrel{x\ \mapsto\ 1/x}{=}\,\,\, &
\int_{1}^{\infty}{2x\,\dd x \over \pars{1 - 4x^{2}}\root{x^{2} - 1}}
\,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
\int_{1}^{\infty}{\dd x \over \pars{1 - 4x}\root{x - 1}}
\\[5mm] \stackrel{x\ =\ t^{2} + 1}{=}\,\,\, &
2\int_{0}^{\infty}{\dd t \over -3 - 4t^{2}} =
\bbx{\ds{-\,{\root{3} \over 6}\,\pi}}
\end{align}
A: If you set $x=\sin\arctan t=\frac{t}{\sqrt{t^2+1}}$ you get an elementary integral:
$$ \int_{-\infty}^{+\infty}\frac{dt}{t^2-4(1+t^2)} = \color{red}{-\frac{\pi}{2\sqrt{3}}}.$$
