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:
A2x−3+B2x+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(2×32)+3/(2x−3)+12×(−32)−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 (16)(12x−3−12x+3). So the integral is now
(16)∫12x−3−12x+3dx
=(16)((12)ln|2x−3|−(12)ln|2x+3|)+C
=112ln∣∣∣2x−32x+3∣∣∣+C
If you don't like the cover-up rule, then write:
14x2−9≡A2x−3+B2x+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=6AandA=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=32 and then x=−32 for the two different values, a trick which is equivalent to the cover-up rule.
