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.
 A: It is generally a helpful principle to get rid of the square roots. In order to do that you need to substitute $x$ with a trigonometric function for which $x^2 - 1$ is also a square. A little thought will show you that $x = \sec \theta$ will do the job.
So $$ \int_1^{\sqrt{2}} \dfrac{dx}{1+ \sqrt{x^2 - 1}} = \int_0^{\frac{\pi}{4}} \dfrac{\sec \theta \tan \theta d\theta}{1+ \sqrt{\sec^2 \theta - 1}}. $$
Can you finish it now?
A: 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
$$\int_1^{\sqrt{2}} \frac{dx}{1+\sqrt{x^2-1}} = \left ( 1-\frac1{\sqrt{2}}\right ) \log{\left (1+\sqrt{2}\right )}$$
A: Picking up on the answer from Nubres: we have
$$I=\int_0^{\pi/4} \frac{\sec\theta\tan\theta\,d\theta}{1+\sqrt{\sec^2\theta-1}}
  =\int_0^{\pi/4} \frac{\tan\theta\,d\theta}{\cos\theta+\sin\theta}\ .$$
I assume from comments that user129823 has already got this far, however it's true that this is still rather an annoying integral.  However any rational function of $\sin\theta$ and $\cos\theta$ can "in principle" (famous last words) be integrated by using the substitution
$$z=\tan\frac{\theta}{2}\ .$$
If you have not studied this yet, have a look here.  In this case, as often happens, the algebra is pretty messy, but if I have done it accurately it leads to
$$I=\int_0^{\tan(\pi/8)} \frac{4z\,dz}{(1-z)(1+z)(\sqrt2-1+z)(\sqrt2+1-z)}$$
which can now be expanded in partial fractions and integrated to give a sum of logarithmic terms.
