How do you integrate #int 1/(4x^2 - 9)# using partial fractions?

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Answer 1 · 2017-02-05T00:16:09

#int (dx)/(4x^2 -9) = 1/12ln abs((2x-3)/(2x+3))+C#

Factorize the denominator:

#4x^2 -9 = (2x)^2 -3^2 = (2x-3)(2x+3)#

Now develop the integrand in partial fractions:

#1/(4x^2 -9) = A/(2x-3)+B/(2x+3)#

#1/(4x^2 -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) =1#

#2Ax+3A+2Bx -3B = 1#

#x(2A+2B) +(3A-3B) = 1#

Equating the coefficient with the same degree in #x#:

#{(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#