Evaluating the indefinite integral $\int\log\!\left(x+\sqrt{x^2-1}\right)\!dx$

Integrate by parts: $$ x\log(x+\sqrt{x^2-1})-\int x\frac{1+\dfrac{x}{\sqrt{x^2-1}}}{x+\sqrt{x^2-1}}\,dx $$

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Now, where did I see $\log(x+\sqrt{x^2-1})$ again? Set $x+\sqrt{x^2-1}=e^t$, so $$ x^2-1=e^{2t}-2xe^t+x^2 $$ or $$ 2xe^t=e^{2t}+1 $$ and $$ x=\cosh t $$ So a good substitution could be this one, wouldn't it? The integral becomes $$ \int t\sinh t\,dt=t\cosh t-\int \cosh t\,dt=t\cosh t-\sinh t $$ and now it's just back substitution.

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.