The denominator is the difference of the squares of 2x and 3, so it easily factorizes as (2x-3)(2x+3) giving the integrand as 1/((2x-3)(2x+3)).
So the fraction is:
A/(2x-3)+B/(2x+3) where A and B have to be determined by you`r method of choice (e.g. cover up rule, or equate coefficients, or susbstitute small values). Here is the cover-up rule:
1/((2xx3/2)+3)/(2x-3)+1/(2xx(-3/2)-3)/(2x+3)
(To get A you look at what value of x makes the expression under A zero, cover up the 2x-3 and replace x with that value in whatever is still visible. Similarly B.)
This simplifies to (1/6)(1/(2x-3)-1/(2x+3)). So the integral is now
(1/6)int 1/(2x-3)-1/(2x+3)dx
=(1/6)((1/2)ln|2x-3|-(1/2)ln|2x+3|)+C
=1/12ln|(2x-3)/(2x+3)|+C
If you don't like the cover-up rule, then write:
1/(4x^2-9)-=A/(2x-3)+B/(2x+3)
which gives:
1-=(2x+3)A+(2x-3)B
which, upon collecting similar powers of x, gives
0x+1-=(2A+2B)x+(3A-2B).
Hence, equating equal powers of x you get:
0=2A+2B hence A=-B and
1=3A-3B hence 1=6A and A=1/6, B=-1/6#.
If you don't like equating similar powers, substitute any two different values for x in the identity 1-=(2x+3)A+(2x-3)B and solve the resulting simultaneous equations. If you don't like solving simultaneous equations, choose x=3/2 and then x=-3/2 for the two different values, a trick which is equivalent to the cover-up rule.
