Question

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

Answer 1

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

Explanation:

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`