Evaluating the indefinite integral $\int\log\!\left(x+\sqrt{x^2-1}\right)\!dx$ I came across the following integral, and I don't know how to solve it.
$$
\int\log\left(x+\sqrt{x^2-1}\right)dx
$$
I tried the "obvious" substitution of $x=\sec\theta$, which gives you:
$$
\int\tan\theta\sec\theta\log\left(\tan\theta+\sec\theta\right)d\theta
$$
However, this doesn't simplify it much, in the sense that I have no idea how to solve this now! Take out the common factor of $\sec$ from the log, or perhaps do $u=\tan\theta+\sec\theta$, but both of these lead to dead ends (at least for me).
In case you are wondering, Wolfram|Alpha claims the answer is:
$$x\log\left(x+\sqrt{x^2-1}\right)-\sqrt{x^2-1}+c$$

Computing directly shows that the integrand, $$\log\!\left(x + \sqrt{x^2 - 1}\right),$$ is the inverse of (the restriction to $[0, \infty)$ of) the hyperbolic cosine function $$\cosh u := \frac{e^u + e^{-u}}{2};$$ because of this, the integrand here is usually denoted $$\text{arcosh x}.$$
This suggests that we may proceed analogously to the usual derivation of the antiderivatives of inverse trigonometric functions: Substituting $x = \cosh u$ gives
$$\int \text{arcosh } x \,dx = \int \text{arcosh} (\cosh u) \,d(\cosh u) = \int u \sinh u \,du.$$
Applying integration by parts with $v = u$, $dw = \sinh u \,du$ gives that this is
$$u \cosh u - \int \cosh u\,du = u \cosh u - \sinh u + C,$$ and reverse-substituting to write this in terms of $x$ yields
$$\text{arcosh } x \cosh (\text{arcosh } x) - \sinh(\text{arcosh } x) + C.$$
Substituting $u = \text{arcosh x}$ in the familiar identity $$\cosh^2 u = \sinh^2 u + 1,$$ simplifying, rearranging, and using that $\text{arcosh}$ is nonnegative (or, alternatively, appealing to the hyperbolic analogue of a reference triangle) gives the identity
$$\sinh (\text{arcosh } x) = \sqrt{x^2 - 1}.$$ Substituing in the above expression gives the antiderivative,
$$\color{#bf0000}{\int \text{arcosh } x \,dx = x \,\text{arcosh } x - \sqrt{x^2 - 1} + C},$$ which in particular agrees with the result given by WolframAlpha.
