It is not difficult to verify that $$ \frac{\mathrm d}{\mathrm dx} \left[ \log\Big(x+\sqrt{x^2+1}\Big) \right] = \frac{1}{\sqrt{1+x^2}} $$ for $x\geq 0$, say.
How would one calculate the indefinite integral $$ \int \frac{1}{\sqrt{1+x^2}} \ \!\mathrm dx$$ without knowing this? I have tried many of the usual tricks, without success.
The title of the question is chosen because Mathematica outputs $\text{Arcsinh}(x)+C$ as the answer.
One would use Euler's substitution. Also on PlanetMath.
Alternatively, one could use $x = \sinh(t)$, because $1+x^2 = 1+\sinh^2(t) = \cosh^2(t)$, and because $\mathrm{d}x = \sinh^\prime(t) \mathrm{d}t = \cosh(t) \mathrm{d}t$. Therefore, using that $\cosh(t) >0 $ for real $t$:
$$ \int \frac{\mathrm{d} x}{\sqrt{1+x^2}} = \int \frac{\cosh(t)}{\sqrt{\cosh^2(t)}} \mathrm{d} t = \int \mathrm{d} t = t + C $$ Substituting back $\int \frac{\mathrm{d} x}{\sqrt{1+x^2}} = \operatorname{arcsinh}(x) + C$
You can do a substitution for $x=\tan\theta$ and $\mathrm dx=\sec^2\theta\,\mathrm d\theta$ to get
$$ \int\frac{\sec^2\theta}{\sqrt{1+\tan^2\theta}}\,\mathrm d\theta. $$
Then use that $1+\tan^2\theta=\sec^2\theta$ to get the integral
$$ \int\sec\theta\,\mathrm d\theta. $$
EDIT: In response to the comment by Sasha: This is one of the standard integrals in a calculus class. I usually just derive it in class and have students memorize it along with the other trigonometric functions. Just multiply top and bottom by $\sec\theta + \tan\theta$ to get
$$ \int \frac{\sec^2\theta + \tan\theta\sec\theta}{\sec\theta+\tan\theta}\mathrm d\theta. $$
Then $u=\sec\theta + \tan\theta$, $\mathrm du=\sec\theta\tan\theta + \sec^2\theta\;\mathrm d\theta.$ The answer is then $\ln|\sec\theta+\tan\theta|+C$. Putting this in terms of $x$ one gets $\ln|\sqrt{1+x^2}+x|+C$.
