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

Feb 5, 2017

$\int \frac{\mathrm{dx}}{4 {x}^{2} - 9} = \frac{1}{12} \ln \left\mid \frac{2 x - 3}{2 x + 3} \right\mid + C$

#### Explanation:

Factorize the denominator:

$4 {x}^{2} - 9 = {\left(2 x\right)}^{2} - {3}^{2} = \left(2 x - 3\right) \left(2 x + 3\right)$

Now develop the integrand in partial fractions:

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

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

As the denominators are equal then also the numerators must be equal for the equation to be satisfied:

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

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

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

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

$\left\{\begin{matrix}2 A + 2 B = 0 \\ 3 A - 3 B = 1\end{matrix}\right.$

From the first we have:

$A = - B$

and substituting this in the second:

$6 A = 1$

and finally:

$\left\{\begin{matrix}A = \frac{1}{6} \\ B = - \frac{1}{6}\end{matrix}\right.$

So:

$\frac{1}{4 {x}^{2} - 9} = \frac{1}{6} \left(\frac{1}{2 x - 3}\right) - \frac{1}{6} \left(\frac{1}{2 x + 3}\right)$

Now solving the integral:

$\int \frac{\mathrm{dx}}{4 {x}^{2} - 9} = \frac{1}{6} \int \frac{\mathrm{dx}}{2 x - 3} - \frac{1}{6} \int \frac{\mathrm{dx}}{2 x + 3}$

$\int \frac{\mathrm{dx}}{4 {x}^{2} - 9} = \frac{1}{12} \int \frac{d \left(2 x - 3\right)}{2 x - 3} - \frac{1}{12} \int \frac{d \left(2 x + 3\right)}{2 x + 3}$

$\int \frac{\mathrm{dx}}{4 {x}^{2} - 9} = \frac{1}{12} \ln \left\mid 2 x - 3 \right\mid - \frac{1}{12} \ln \left\mid 2 x + 3 \right\mid + C$

Using the properties of logarithms we can also write it as:

$\int \frac{\mathrm{dx}}{4 {x}^{2} - 9} = \frac{1}{12} \ln \left\mid \frac{2 x - 3}{2 x + 3} \right\mid + C$