Substitute #x^2+5=u^2#. Then, #2xdx=2udu# and thus # dx/x =(udu)/x^2=(udu)/(u^2-5)#.
So, our integral becomes
#int sqrt(x^2+5)/x dx = int u (udu)/(u^2-5)# #qquad = int (u^2-5+5)/(u^2-5) = int (1+5/(u^2-5))du# #qquad = u+5 int (du)/(u^2-(sqrt5)^2)= u + 5/(2sqrt5) ln((u-sqrt5)/(u+sqrt5))# #qquad = sqrt(x^2+5) +5/(2sqrt5) ln((sqrt(x^2+5)-sqrt5)/(sqrt(x^2+5)+sqrt5))#