Let p=(x^2-1)^(1/2); p^2=x^2-1; x^2=1+p^2; dp=xdx/sqrt(x^2-1),
so dx/(sqrt(x^2-1)=dp/x and dx/(xsqrt(x^2-1))=dp/x^2=dp/(1+p^2).
When x=sqrt(2), p=1 and when x=2, p=sqrt(3).
Let p=tan(y); 1+p^2=sec^2(y) and dp=sec^2(y)dy; dp/(1+p^2)=sec^2(y)dy/sec^2(y)=dy.
When 1=p=tan(y), y=(pi)/4 and when sqrt(3)=p=tan(y), y=(pi)/3. So the integral is y[(pi)/4,(pi)/3]=(pi)(1/3-1/4)=(pi)/12.