$\int \frac{dx}{9x^2 - 16}$ is equal to

The denominator is $9x^2 - 16$.

Notice that:

$9x^2 = (3x)^2$

$16 = 4^2$

So $9x^2 - 16 = (3x)^2 - (4)^2 = (3x - 4)(3x + 4)$

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The integral becomes:

$\int \frac{dx}{(3x-4)(3x+4)}$

Using partial fractions:

$\frac{1}{(3x-4)(3x+4)} = \frac{A}{3x-4} + \frac{B}{3x+4}$

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Multiply both sides by $(3x-4)(3x+4)$:

$1 = A(3x+4) + B(3x-4)$

Put $3x = 4$:

$1 = A(4+4)$

$1 = 8A$

$A = \frac{1}{8}$

Put $3x = -4$:

$1 = B(-4-4)$

$1 = -8B$

$B = -\frac{1}{8}$

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The integral becomes:

$\int \frac{dx}{9x^2 - 16} = \int \left(\frac{1/8}{3x-4} - \frac{1/8}{3x+4}\right)dx$

$= \frac{1}{8}\int \frac{dx}{3x-4} - \frac{1}{8}\int \frac{dx}{3x+4}$

$= \frac{1}{8} \cdot \frac{1}{3}\ln|3x-4| - \frac{1}{8} \cdot \frac{1}{3}\ln|3x+4| + C$

$= \frac{1}{24}\ln|3x-4| - \frac{1}{24}\ln|3x+4| + C$

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Using the logarithm property $\ln a - \ln b = \ln\left(\frac{a}{b}\right)$:

$= \frac{1}{24}\left[\ln|3x-4| - \ln|3x+4|\right] + C$

$= \frac{1}{24}\ln\left|\frac{3x-4}{3x+4}\right| + C$

Therefore, $\int \frac{dx}{9x^2 - 16} = \frac{1}{24}\ln\left|\frac{3x-4}{3x+4}\right| + C$