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

$\frac{1}{24}\log_e\left|\frac{3x-4}{3x+4}\right|+C$, Where C is constant of integration

The correct answer is Option (4) → $\frac{1}{24}\log_e\left|\frac{3x-4}{3x+4}\right|+C$, Where C is constant of integration

Evaluate

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

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

Use partial fractions

$\frac{1}{9x^2-16}=\frac{A}{3x-4}+\frac{B}{3x+4}$

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

Putting $x=\frac{4}{3}$

$1=A(8)$, so $A=\frac{1}{8}$

Putting $x=-\frac{4}{3}$

$1=B(-8)$, so $B=-\frac{1}{8}$

Hence

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

$=\frac{1}{8}\left[\frac{1}{3}\log|3x-4|-\frac{1}{3}\log|3x+4|\right]+C$

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

The value of the integral is $\frac{1}{24}\log\left|\frac{3x-4}{3x+4}\right|+C$.