Help evaluating $ \int \sqrt{{x}^{2} + 3} \; dx $ Can you help me evaluating the following indefinite integral?
$$
\int \sqrt{{x}^{2} + 3} \; dx
$$
Please, don't give a full solution, just some hint on which method to use...
** UPDATE **
Thank you very much to everybody for the useful comments and suggestions. I'm sorry for the delay with my reply, unfortunately do some mathematics as an hobby and often don't have time to work at it.
I tried to take Lucian suggestion on board and use trigonometric substitution as follows.
$$ x = \sqrt{3} \tan\theta $$
and
$$ \int \sqrt{{x}^{2} + 3} \; dx = \int \sqrt{{3\tan}^{2}\theta + 3} \; \sqrt{3}\sec^{2}\theta \; d\theta = \int \sqrt{{3\sec}^{2}\theta} \; \sqrt{3}\sec^{2}\theta \; d\theta = \int \sqrt{3}\sec\theta \; \sqrt{3}\sec^{2}\theta \; d\theta = 3 \int \sec^{3}\theta \; d\theta $$
which (according to common integral tables) is equal to
$$ 3 \left[ \frac{1}{2} \sec\theta \tan\theta + \frac{1}{2} \ln\left| \sec\theta + \tan\theta \right| + C \right] $$
Problem arises when I try to substitute back the variable from $ \theta $ to $ x $ because I know $ \tan\theta = \frac{x}{\sqrt{3}} $ but I don't know how to substitute back $ \sec\theta $, so basically I stopped here:
$$
\frac{3}{2}\sec\theta\;\frac{x}{\sqrt{3}} + \frac{3}{2}\ln\left| \sec\theta + \frac{x}{\sqrt{3}} \right| + 3C 
$$
It looks close to the final answer but still not there...any suggestions?

Substitution $\sqrt{x^2+3}=u-x$ will work
$\sqrt{x^2+3}=u-x\\ x^2 +3 = u^2 - 2ux + x^2\\3=u^2-2ux \\ 2ux = u^2-3 \\ x = \frac{u^2-3}{2u} \\ u - x = u - \frac{u^2-3}{2u}\\u-x=\frac{u^2+3}{2u} \\ 
dx = \frac{2u\cdot2u-2(u^2-3)}{4u^2}du \\ dx = \frac{2u^2+6}{4u^2}du \\ dx = \frac{u^2+3}{2u^2}du \\ \int{\frac{u^2+3}{2u^2}\cdot\frac{u^2+3}{2u}du} \\ \frac{1}{4}\int{\frac{u^4+6u^2+9}{u^3}du}\\
\frac{1}{4}\left(\frac{1}{2}u^2-\frac{9}{2}\frac{1}{u^2}\right)+\frac{3}{2}\ln{|u|}+C\\\frac{1}{2}(\frac{u^4-9}{4u^2})+\frac{3}{2}\ln{|u|}+C\\
\frac{1}{2}\frac{u^2-3}{2u} \cdot \frac{u^2+3}{2u}+\frac{3}{2}\ln{|u|}+C \\
\frac{1}{2}x\sqrt{x^2+3} + \frac{3}{2}\ln{|x+\sqrt{x^2+3}|}+C
$
$x=\sqrt{3}\tan{\theta}\\
\frac{x}{\sqrt{3}} = \tan{\theta}
$

Recall identity $1+\tan^2{\theta}=\sec^2(\theta)$
$\frac{x^2}{3}=\tan^2{\theta}\\
1+\frac{x^2}{3} = 1+ \tan^2{\theta} \\
1+\frac{x^2}{3} = \sec^2{\theta}\\
\sqrt{\frac{3+x^2}{3}} = \sec{\theta}
$

but you must be sure that secant is nonnegative 
In my opinion subsitution which i showed is better for this integral  because after this substitution we only need to integrate power function
