Indefinite integral $\int{\frac{dx}{x^2+2}}$

Hint:

$$x^2+2 = 2\left(\frac{x^2}{\sqrt{2}^2}+1\right)$$

Hint: take $t=\frac{x}{\sqrt 2}$.

I find it much more versatile when encountering a denominator of the form $x^2 + a^2$, rather than only having learned what to do when $a = 1$, I use the fact that : $$\int \dfrac{dx}{x^2 + a^2} = \dfrac 1a\arctan\left(\frac x{a}\right) + C$$

Why? $$\frac{dx}{x^2+a^2} = \frac{dx}{a^2 \left(\frac{x^2}{a^2} + 1\right)} =\frac{dx}{a^2\left(\left(\frac{x}{a}\right)^2+1\right)} = \dfrac 1a\cdot\frac{(1/a) \,dx}{\left(\left(\frac{x}{a}\right)^2+1\right)} = \frac{1}{a}\cdot\frac{du}{u^2+1}, \;\;u = \frac xa$$

Applying this fact to your integral is rather straightforward then:

$$\int{\frac{dx}{x^2+2}} = \int\frac{dx}{x^2 + \left(\sqrt 2\right)^2} = \frac 1{\sqrt 2} \arctan\left(\frac x{\sqrt 2}\right) + C$$