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?
 A: One may recall that
$$
\cosh^2 u -\sinh^2 u=1 \quad \text{or}\quad \cosh^2 u =1+\sinh^2 u,
$$
then you may try the change of variable 
$$x=\sqrt{3}\sinh u, \quad dx=\sqrt{3}\cosh u\: du,$$ giving
$$
\begin{align}
\int\sqrt{x^2+3}\:dx &=\int\sqrt{3\:(\sinh^2 u+1)}\times \sqrt{3}\cosh u\: du\\\\
 &=3\int\cosh^2 u \:du\\\\
&=\frac{3}2\int\left(1+\cosh(2 u)\right) \:du.
\end{align}
$$ Hoping you can take it from here.
A: 
Please, don't give a full solution, just some hint on which method to use...

Hint: Let $x^2=3\tan^2t$, and use the fact that $\tan't=1+\tan^2t=\dfrac1{\cos^2t}$
A: You have modified your question and now you want to determine the substitution for $\sec\theta$. So, here we go:
Given: 
$$ \tan\theta = \frac{x}{\sqrt{3}} $$
knowing that $\tan\theta$ is positive for positive $x$, squaring both sides:
$$ tan^2\theta = \frac{x^2}{3} $$
$$ 1+ tan^2\theta = 1+ \frac{x^2}{3} $$
$$ 1+ \frac{sin^2\theta}{cos^2\theta} = 1+ \frac{x^2}{3} $$
$$ \frac{cos^2\theta + sin^2\theta}{cos^2\theta} = 1+ \frac{x^2}{3} $$
and since $ cos^2\theta + sin^2\theta = 1 $, we have:
$$ \frac{1}{cos^2\theta} = 1+ \frac{x^2}{3} $$
$$ \sec^2\theta = 1+ \frac{x^2}{3} $$
$$ \sec\theta = \sqrt{1+ \frac{x^2}{3}} $$
This is your substitution. One would argue that how did I remove squares from both sides, isn't that illegal? Well, not here - since I was the one who squared the sides in the first place, I know that the positive square root is the correct one, as the negative one was the result of me squaring the equation and creating an additional root of the equation. The positive root is consistent with the equation I started with, hence it's the one we are looking for.
A: Just a hint. A possible approach would be to consider using the formula for the integral of an irrational function like this: $r={\sqrt  {a^{2}+x^{2}}}$ so that $\int r\;dx={\frac  {1}{2}}\left(xr+a^{2}\,\ln \left({x+r}\right)\right) + C$. 
For restricted x values it can be represented as an expression with $sinh$.
A: 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
