Find the integral: $\displaystyle \int \frac{dx}{x^2 - 16}$
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{1}{8} \log \left| \frac{x - 4}{x + 4} \right| + C$
We have $\int \frac{dx}{x^2 - 16} = \int \frac{dx}{x^2 - 4^2}$
$= \frac{1}{8} \log \left| \frac{x - 4}{x + 4} \right| + C$