<s>[INST] Integration with substitution of trig: $\int\frac1{(x^2-4)^{3/2}}dx$ $$\int\frac1{(x^2-4)^{3/2}}dx $$
I'm unsure how to go about this integral after subbing $x=2\cosh(\theta)$; to the power of $3/2$ is confusing me when trying to simplify.
Thanks.
 [/INST] Let $u^2x^2=x^2-4$. Then $x=\pm\sqrt{\frac{-4}{u^2-1}}$, $dx=\pm(-4)\sqrt{\frac{u^2-1}{-16}}\left(\frac{-2u}{\left(u^2-1\right)^2}\right)\,du$.
$$\int\frac{1}{\left(x^2-4\right)^{3/2}}\, dx=\int\frac{\pm(-4)\sqrt{\frac{u^2-1}{-16}}\left(\frac{-2u}{\left(u^2-1\right)^2}\right)\,du}{u^3\cdot \pm\left(\frac{-4}{u^2-1}\right)\sqrt{\frac{-4}{u^2-1}}}$$
$$=\frac{1}{4}\int\frac{1}{u^2}\, du=\frac{x}{-4\sqrt{x^2-4}}+C$$
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