The indefinite integral $\int{\frac{\mathrm dx}{\sqrt{1+x^2}}}$ I need to solve this integral:  
$$\int{\frac{\mathrm dx}{\sqrt{1+x^2}}}$$
First I thought it was easy, so I tried integration by parts with $g(x)=x$ and $g'(x)=1$:  
$$\int{ \frac{x^2}{(1+x^2)^{\frac{3}{2}} }}\,\mathrm dx $$
But I've made it even more complicated than before, and if I want to solve it again by parts I'll have $g(x)= \frac{x^3}{3}$ , and I will never end integrating.  
How should I solve it?
Edit 
Trying this way: $x= tg(t)$, then I get:  
$$ \int{ \frac{1+tg^2(t)}{ \sqrt{1+ tg^2(t)} } dt}= \int{ \sqrt{ 1 + tg^2(t) } dt } $$
But it doesn't remind me anything, I still can't solve it.
 A: If you are in a mood for miracles, let $w=x+\sqrt{1+x^2}$. 
Then 
$$dw= \left(1+\frac{x}{\sqrt{1+x^2}}\right)\,dx=w \frac{dx}{\sqrt{1+x^2}},$$
and we end up needing to find $\displaystyle\int \frac{dw}{w}$.
A: When you see $1+x^2$ you should without hesitation think "Let $x=\tan\theta$."
If you're comfortable hyperbolic trig functions, then metacompactness's suggestion is better. :)
A: your integral can be solved by
$x=\sinh t$ and $dx=\cosh t\, dt$ which gives
$$\int \frac{\cosh t}{\cosh t} \, dt = \int 1\, dt=t=\sinh^{-1} x = \ln (x+\sqrt{x^2+1})+C,$$
since $\sqrt{1+x^2}=\sqrt{1+\sinh^2 t}=\cosh t$.
A: This one of the remarkable derivative of the usual hyperbolic  function $\mathrm{arsinh}$ and this is the complete list.
A: Just to complete the OP's second suggested solution, by the change of variables $x=\tan{\theta}$ we get:
$$\int \frac{dx}{\sqrt{1+x^2}}=\int{\sqrt{1+\tan^2\theta}d\theta}=\int \sqrt{\frac{\sin^2 \theta + \cos^2\theta}{cos^2 \theta}}d\theta=\int \frac{d\theta}{\cos\theta}=\int \sec\theta \ d\theta$$
Now this is a well known integral; to take it, notice that $\frac{d}{d\theta}\left(\tan\theta + \sec \theta \right)=\sec \theta \left(\tan\theta + \sec \theta \right)$. So we will write the integral as:
$$\int \sec \theta \ d\theta = \int \frac{\sec \theta \left(\tan\theta + \sec \theta \right)}{\left(\tan\theta + \sec \theta \right)}d\theta=\ln \left(\sec \theta + \tan\theta \right)$$
Now we have to change back the variables:
$$\int \frac{dx}{\sqrt{1+x^2}}=\ln \left(\sec \left(\tan^{-1}{x} \right) + \tan\left( \tan^{-1}{x} \right) \right)=\ln \left(\sqrt{1+x^2} + x \right)$$
For the last equation, you may want to draw a right-triangle.
