Integral of $\int^\sqrt2_1\frac{1}{1+\sqrt{x^2 - 1}}dx$ by substitution? In a maths question I have the question:
$$\int^\sqrt2_1\frac{1}{1+\sqrt{x^2 - 1}}dx$$  by substitution?
All other questions have been by trigonometric substitution so I assume that is how to solve.

Another substitution that is useful here is $x=\cosh{t}$; the integral is thus equal to
$$\int_0^{\log{(1+\sqrt{2})}} dt \frac{\sinh{t}}{1+\sinh{t}} = \log{(1+\sqrt{2})} - \int_0^{\log{(1+\sqrt{2})}}  \frac{dt}{1+\sinh{t}}$$
Now use $\sinh{t} = (e^t-e^{-t})/2$ and sub $y=e^t$ to get that the integral on the right is
$$2 \int_1^{\sqrt{2}+1} \frac{dy}{y^2+2 y-1} = 2 \int_1^{\sqrt{2}+1} \frac{dy}{(y+1)^2-2} = 2 \int_2^{2+\sqrt{2}} \frac{du}{u^2-2}$$
This integral is equal to
$$\frac1{\sqrt{2}} \left [\log{\left (\frac{u-\sqrt{2}}{u+\sqrt{2}} \right )} \right ]_2^{2+\sqrt{2}} = \frac1{\sqrt{2}} \log{\left ( 1+\sqrt{2}\right )}$$
Thus the original integral is
$$\int_1^{\sqrt{2}} \frac{dx}{1+\sqrt{x^2-1}} = \left ( 1-\frac1{\sqrt{2}}\right ) \log{\left (1+\sqrt{2}\right )}$$
