Indefinite integral of $\int\sqrt{x^2-1}dx$ How to calculate $\int\sqrt{{x^2}-1}dx$?
I tried to use 1st and 3rd Euler substitution and got the answer, but I suppose there's an trigonometrical replacing that gives the answer analytically.

Integration without substitution
Notice, there is another method without any substitution, using integration by parts, let
$$I=\int\sqrt{x^2-1}\ dx$$$$=\int \sqrt{x^2-1}\cdot 1dx$$
$$I=\sqrt{x^2-1}\int 1\ dx-\int \left(\frac{d}{dx}\left(\sqrt{x^2-1}\right)\cdot \int1 dx\right)dx$$
$$I=\sqrt{x^2-1}(x)-\int \frac{x(2x)}{2\sqrt{x^2-1}}dx$$
$$I=x\sqrt{x^2-1}-\int \frac{(x^2-1)+1}{\sqrt{x^2-1}}dx$$
$$I=x\sqrt{x^2-1}-\int\sqrt{x^2-1} dx+\int \frac{1}{\sqrt{x^2-1}}dx$$
$$I=x\sqrt{x^2-1}-I+\ln|x+\sqrt{x^2-1}|+C$$
$$2I=x\sqrt{x^2-1}+\ln|x+\sqrt{x^2-1}|+C$$
$$I=\color{red}{\frac{1}{2}\left(x\sqrt{x^2-1}+\ln|x+\sqrt{x^2-1}|\right)+C}$$
