$$\int\ln(x+\sqrt{x^2-1})dx$$ To evaluate the integral we can apply integral by parts (using $u=\ln(x+\sqrt{x^2-1})$ and $dv=dx$). but my question is: is there other approach to evaluate the integral faster? I think there might be something because we can write it as:
$$\ln\left(x+\sqrt{x^2-1}\times\dfrac{x-\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\right)=\ln\left(\cfrac1{x-\sqrt{x^2-1}}\right)=-\ln(x-\sqrt{x^2-1})$$ But I don't know how to use this.
Use $x=\cosh t$
$$\int \sinh t \ln(\cosh t + \sinh t) dt = \int t \sinh t dt = t\cosh t - \sinh t$$
then substitute back.
Note that this is an inverse function and hence IBP is standard (see below). As soon as you recognize $$ \ln(x+\sqrt{x^2-1})=\mbox{arcosh}(x) $$ you can simplify the IBP computations, since no computation is really needed:
The following formula holds for any continuous bijective function: $$ \int f^{-1}(x) dx= x f^{-1}(x)- \int x (f^{-1})'(x) dx = x f^{-1}(x)- \int f( f^{-1}(x)) (f^{-1})'(x) dx \\ = x f^{-1}(x)- \int f(u) du $$
[Side note: there is a picture proof of this formula in the Book "Proofs without Words". The proof is basically just observing that $$ \int_{a}^x f(x) dx + \int_{f(a)}^{f(x)} f^{-1}(u) du $$ graphically the signed between the rectangle with corners $(0,0), (0,f(x)), (x,0), (x,f(x))$ and the rectangle with corners $(0,0), (0,f(a)), (a,0), (a,f(a))$ The picture proof can be seen in this answer. ]
Now set $$f(x)= \cosh(x) \\ f^{-1}(x)=\mbox{arcosh}(x)$$ and integrate $\int \cosh(u) du$.
If you accept the above formula as a "trick", then that's it.
Note Whenever when you know the integral of a bijection $f(x)$, the above formula gives you the integral of $f^{-1}$.
Since the formula is just IBP, it explains why $\int \ln(x) dx, \int \arcsin(x) ds$.... are always easy to calculate by parts.
Note $$\int\ln\left(x+\sqrt{x^2-1}\right)dx=\int \cosh^{-1}xdx \overset{IBP}=x \cosh^{-1}x -\sqrt{x^2-1}$$
