I know how to solve the integral (set $x = a\sec(\theta)$ then $dx = a\sec(\theta)\tan(\theta)\,d\theta$ where $0 < \theta < \pi/2$ in order to have a one-to-one function), anyway, the problem specifies that $a > 0$ but I don't see how this changes anything or affects the substitution because $x = a\sec(\theta)$ still remains a one-to-one function, correct? I may be wrong but by seeing the graph of asec(theta) it appears that the sign of a does not change the fact that sec(theta) is one to one. Does it affect something else?
Then we could also use the substitution of $x = a\cosh(t)$ then $dx = a\sinh(t)\,dt$ but the problem then specifies that $x > 0$. Why these restrictions?