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

Divide the denominator by two:

The integrand is now developed in partial fractions:

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

Starting with the first, we have:

and using this in place of it in the second:

and lastly:

So:

Currently resolving the integral:

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

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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.

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