# 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?

• Replace $x$ with $\sqrt{3} \sinh u$. – Jack D'Aurizio Mar 29 '15 at 15:18
• or Integrate by parts – Suhail Mar 29 '15 at 15:26
• @suhail, what do you after the first step? – abel Mar 29 '15 at 15:29
• @abel See my answer. I hope that it is useful – Mark Viola Mar 29 '15 at 15:56
• @Jashin I really would like to help, so please let me know how I can improve my answer. I just want to give you the best answer I can. – Mark Viola Mar 29 '15 at 15:59

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.
Hint: Let $x^2=3\tan^2t$, and use the fact that $\tan't=1+\tan^2t=\dfrac1{\cos^2t}$