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

Question

Samantha Adkison · Accepted Answer

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

Divide the denominator by two:

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

The integrand is now developed 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))#

For the equation to be satisfied, both the denominators and the numerators must be equal:

#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):}#

Starting with the first, we have:

#A=-B#

and using this in place of it in the second:

#6A=1#

and lastly:

#{(A=1/6),(B=-1/6):}#

So:

#1/(4x^2 -9) = 1/6 (1/(2x-3))-1/6 (1/(2x+3))#

Currently resolving 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#

We can also express it as follows by using the logarithm properties:

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

Cameron Alexander · Answer

undefined To integrate $ \int \frac{1}{4x^2 - 9} $ using partial fractions, first factor the denominator:

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

Now, express $ \frac{1}{4x^2 - 9} $ as a sum of partial fractions:

$$ \frac{1}{4x^2 - 9} = \frac{A}{2x - 3} + \frac{B}{2x + 3} $$

To solve for $ A $ and $ B $, multiply both sides by $ 4x^2 - 9 $ to clear the fractions:

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

Now, substitute suitable values for $ x $ to solve for $ A $ and $ B $. For example, setting $ x = \frac{3}{2} $ eliminates $ B $, and setting $ x = -\frac{3}{2} $ eliminates $ A $.

$$ 1 = A(3) + 0 \Rightarrow A = \frac{1}{3} $$
$$ 1 = 0 + B(-3) \Rightarrow B = -\frac{1}{3} $$

Now that you have found the values of $ A $ and $ B $, rewrite the integral with the partial fractions:

$$ \int \frac{1}{4x^2 - 9} \, dx = \int \frac{\frac{1}{3}}{2x - 3} - \frac{\frac{1}{3}}{2x + 3} \, dx $$

Now integrate each term separately:

$$ = \frac{1}{3} \int \frac{1}{2x - 3} \, dx - \frac{1}{3} \int \frac{1}{2x + 3} \, dx $$

$$ = \frac{1}{3} \ln|2x - 3| - \frac{1}{3} \ln|2x + 3| + C $$

So, $ \int \frac{1}{4x^2 - 9} \, dx = \frac{1}{3} \ln|2x - 3| - \frac{1}{3} \ln|2x + 3| + C $, where $ C $ is the constant of integration.