Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I cannot manage to solve this integral:


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?

share|cite|improve this question
$x^2 + 2 = 2(t^2 + 1)$ <- what's $t$ then? – Daniel Fischer Jul 13 '13 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 $u=\frac{x}{a}$ – user84413 Jul 13 '13 at 20:51
I'm pretty sure the very same integral has been solved here in 17 different threads by now. – NikolajK Jul 13 '13 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. – Ramy Al Zuhouri Jul 13 '13 at 21:46
@RamyAlZuhouri: One option is having it solved by Wolphram alpha via "Integrate[a/(b+x^2),x]" and reverse engineer. – NikolajK Jul 13 '13 at 22:04
up vote 8 down vote accepted


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

share|cite|improve this answer

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

share|cite|improve this answer

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:

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

share|cite|improve this answer
In your first equation, $a^2$ goes to $\sqrt{a}$. But in the second equation, $2$ goes to $\sqrt{2}$? – DJohnM Jul 13 '13 at 22:11

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

share|cite|improve this answer


Now take $u=\frac{x}{\sqrt{2}}$ and $du=\frac{1}{\sqrt{2}}dx$ so that we get


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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.