Factorize the denominator:
4x^2 -9 = (2x)^2 -3^2 = (2x-3)(2x+3)4x2−9=(2x)2−32=(2x−3)(2x+3)
Now develop the integrand in partial fractions:
1/(4x^2 -9) = A/(2x-3)+B/(2x+3)14x2−9=A2x−3+B2x+3
1/(4x^2 -9) = (A(2x+3)+B(2x-3))/((2x+3)(2x-3))14x2−9=A(2x+3)+B(2x−3)(2x+3)(2x−3)
As the denominators are equal then also the numerators must be equal for the equation to be satisfied:
A(2x+3)+B(2x-3) =1A(2x+3)+B(2x−3)=1
2Ax+3A+2Bx -3B = 12Ax+3A+2Bx−3B=1
x(2A+2B) +(3A-3B) = 1x(2A+2B)+(3A−3B)=1
Equating the coefficient with the same degree in xx:
{(2A+2B = 0),(3A-3B = 1):}
From the first we have:
A=-B
and substituting this in the second:
6A=1
and finally:
{(A=1/6),(B=-1/6):}
So:
1/(4x^2 -9) = 1/6 (1/(2x-3))-1/6 (1/(2x+3))
Now solving the integral:
int (dx)/(4x^2 -9) = 1/6int (dx)/(2x-3)-1/6 int(dx)/(2x+3)
int (dx)/(4x^2 -9) = 1/12int (d(2x-3))/(2x-3)-1/12 int(d(2x+3))/(2x+3)
int (dx)/(4x^2 -9) = 1/12ln abs(2x-3)-1/12 ln abs(2x+3)+C
Using the properties of logarithms we can also write it as:
int (dx)/(4x^2 -9) = 1/12ln abs((2x-3)/(2x+3))+C
