Indefinite integral $\int{\frac{dx}{x^2+2}}$ I cannot manage to solve this integral:  
$$\int{\frac{dx}{x^2+2}}$$
The problem is the $2$ at denominator, I am trying to decompose it in something like $\int{\frac{dt}{t^2+1}}$:  
$$t^2+1 = x^2 +2$$
$$\int{\frac{dt}{2 \cdot \sqrt{t^2-1} \cdot (t^2+1)}}$$
But it's even harder than the original one. I also cannot try partial fraction decomposition because the polynomial has no roots. Ho to go on?
 A: Hint:
$$x^2+2 = 2\left(\frac{x^2}{\sqrt{2}^2}+1\right)$$
A: Hint: take $t=\frac{x}{\sqrt 2}$.
A: 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:
$$\int{\frac{dx}{x^2+2}} = \int\frac{dx}{x^2 + \left(\sqrt 2\right)^2} = \frac 1{\sqrt 2} \arctan\left(\frac x{\sqrt 2}\right) + C$$
A: $$
\frac{dx}{x^2+2} = \frac{dx}{2\left(\frac{x^2}{2} + 1\right)} =\frac{dx}{2\left(\left(\frac{x}{\sqrt{2}}\right)^2+1\right)} = \frac{dx/\sqrt{2}}{\sqrt{2}\left(\left(\frac{x}{\sqrt{2}}\right)^2+1\right)} = \frac{1}{\sqrt{2}}\cdot\frac{du}{u^2+1}
$$
A: $$\int\frac{dx}{x^2+2}=\frac{1}{2}\int\frac{dx}{(\frac{x}{\sqrt{2}})^2+1}$$
Now take $u=\frac{x}{\sqrt{2}}$ and $du=\frac{1}{\sqrt{2}}dx$ so that we get
$$\frac{\sqrt{2}}{2}\int\frac{du}{u^{2}+1}.$$
This last integral can be evaluated since $\int\frac{du}{u^{2}+1}=\arctan(u)+C$ where C is a constant. This means the integral we were considering is
$\frac{\sqrt{2}}{2}\arctan(u)+D$ where D is an arbitrary constant.
