This can be done with partial fractions, but this will be done with a trigonometric substitution.
First, we need to add dx to the integral:
I=int1/(x^2-4)dx
Recall that sec^2theta-1=tan^2theta.
Because of that identity, let x=2sectheta.
Importantly, this implies that x^2-4=4sec^2theta-4=4(sec^2theta-1)=4tan^2theta.
Furthermore, dx=2secthetatanthetad theta.
Then:
I=int(2secthetatantheta)/(4tan^2theta)d theta=1/2intsectheta/tanthetad theta=1/2intcscthetad theta
Which is a commonly known integral:
I=-1/2lnabs(csctheta+cottheta)
From our original substitution x=2sectheta we see that sectheta=x/2. This is a right triangle where 2 is the side adjacent to theta, x is the hypotenuse and then the side opposite theta is sqrt(4-x^2).
Then, csctheta=x/sqrt(4-x^2) and cottheta=2/sqrt(4-x^2). So:
I=-1/2lnabs(x/sqrt(4-x^2)+2/sqrt(4-x^2))=1/2lnabs(sqrt(4-x^2)/(x+2))
Expanding further:
I=1/4lnabs(4-x^2)-1/2lnabs(x+2)=1/4lnabs(2+x)+1/4lnabs(2-x)-1/2lnabs(x+2)
I=1/4lnabs(2-x)-1/4lnabs(x+2)
We can combine these and switch the order of 2-x to x-2 since we're working with absolute values:
I=1/4lnabs((x-2)/(x+2))+C
This would be much quicker if done with partial fractions.