TITLE: Calculus Question: $\int\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}+x-1}dx$
QUESTION [5 upvotes]: How to evaluate integral $$\int\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}+x-1}dx?$$
I tried substitution $u^2=x^2-1$ and $u=\sqrt{x^2-1}+x$ but it turns out too complicated. Could anyone here help me to evaluate the integral? Thanks in advance.

REPLY [2 votes]: Could anyone here help me to evaluate the integral? Thanks in advance.

As a general rule, whenever dealing with expressions containing $\sqrt{x^2\pm1}$, the natural substitution is $x=\cosh t$, $($for $-)$, or $x=\sinh t$, $($for $+)$, since $\cosh^2t-\sinh^2t=1.~$ Then, after using the fact that $\cosh t\pm\sinh t=e^{\large\pm t}$, together with $\cosh't=\sinh t$ and $\sinh't=\cosh t,~$ let $e^t=u$, and employ partial fraction decomposition.