How do you do partial fraction $\frac2{x^2 - 8}$? How do you do partial fraction on $\frac2{x^2 - 8}$ ? Or are there other method of doing? I tried trig substitution but could not get the answer.

It's very simple:
$$\frac{2}{x^2-8} = \frac{2}{(x+2\sqrt 2)(x-2\sqrt 2)} = \frac{\frac{1}{4} \sqrt 2}{x-2\sqrt 2}-\frac{\frac{1}{4} \sqrt 2}{x+2\sqrt 2}$$
Thus, for example, $$\int \frac{2}{x^2-8} dx = \frac{1}{4} \sqrt 2 \int \frac{dx}{x-2\sqrt 2} - \frac{1}{4} \sqrt 2 \int \frac{dx}{x+2\sqrt 2}$$
