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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?

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$x^2 + 2 = 2(t^2 + 1)$ <- 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

5 Answers 5

up vote 8 down vote accepted


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

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

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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:

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

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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

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

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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.

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