If our integral involves a root in the form of #sqrt(x^2-a^2),# we're best off using the substitution
#x=asectheta.# Here, we see #a^2=4, a=2,# so we use the substitution
#x=2sectheta#
#dx=2secthetatanthetad theta#
Apply to the integral #intdx/sqrt(x^2-4),# yielding
#int(2secthetatantheta)/sqrt(4sec^2theta-4)d theta#
Simplify.
#int(2secthetatantheta)/sqrt(4(sec^2theta-1)d theta#
#int(secthetatantheta)/sqrt(sec^2theta-1)d theta#
Recalling the identity #sec^2theta-1=tan^2theta,# we get
#int(secthetatantheta)/sqrt(tan^2theta)d theta=int(secthetacanceltantheta)/canceltanthetad theta#
So, we end up with the common integral (you should have this memorized)
#intsecthetad theta=ln|sectheta+tantheta|+C#
We need things in terms of #x.# Well, recalling that #x=2sectheta, sectheta=x/2.# We still need to find tangent.
Recalling that #sec^2theta-1=tan^2theta, x^2/4-4/4=tan^2theta, tan^2theta=(x^2-4)/4, tantheta=sqrt(x^2-4)/2#
Thus, we have
#intdx/sqrt(x^2-4)=ln|(x+sqrt(x^2-4))/2|+C#
#ln|(x+sqrt(x^4-4))/2|=ln|x+sqrt(x^2-4)|-ln|2|+C=ln|x+sqrt(x^2-4)|+C#, as we can absorb that logarithm into the constant of integration.
Finally,
#intdx/sqrt(x^2-4)=ln|(x+sqrt(x^2-4))|+C#
