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.
