# 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?

• $x^2 + 2 = 2(t^2 + 1)$ <- what's $t$ then? Commented Jul 13, 2013 at 20:47
• In general, if you have $x^2+a^2$ in the denominator, you can factor out the $a^2$ to get $(a^2)(\frac{x^2}{a^2}+1)$ in the denominator, and then substitute $u=\frac{x}{a}$ Commented Jul 13, 2013 at 20:51
• I'm pretty sure the very same integral has been solved here in 17 different threads by now. Commented Jul 13, 2013 at 21:27
• @NickKidman I tried to search a similar question with google, but the formula is written with mathjax, and probably it isn't recognized by google, I haven't found any useful result. Commented Jul 13, 2013 at 21:46
• @RamyAlZuhouri: One option is having it solved by Wolphram alpha via "Integrate[a/(b+x^2),x]" and reverse engineer. Commented Jul 13, 2013 at 22:04

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:

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

• In your first equation, $a^2$ goes to $\sqrt{a}$. But in the second equation, $2$ goes to $\sqrt{2}$? Commented Jul 13, 2013 at 22:11

$$\frac{dx}{x^2+2} = \frac{dx}{2\left(\frac{x^2}{2} + 1\right)} =\frac{dx}{2\left(\left(\frac{x}{\sqrt{2}}\right)^2+1\right)} = \frac{dx/\sqrt{2}}{\sqrt{2}\left(\left(\frac{x}{\sqrt{2}}\right)^2+1\right)} = \frac{1}{\sqrt{2}}\cdot\frac{du}{u^2+1}$$

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

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.