CUET Preparation Today
Find the integral: $\displaystyle \int \frac{dx}{x^2 - 16}$ |
$\frac{1}{8} \log \left| \frac{x + 4}{x + 3} \right| + C$ $\frac{5}{8} \log \left| \frac{x - 4}{x + 4} \right| + C$ $\frac{1}{8} \log \left| \frac{x - 4}{x + 4} \right| + C$ $\frac{3}{8} \log \left| \frac{x - 4}{x + 4} \right| + C$ |
$\frac{1}{8} \log \left| \frac{x - 4}{x + 4} \right| + C$ |
The correct answer is Option (3) → $\frac{1}{8} \log \left| \frac{x - 4}{x + 4} \right| + C$ We have $\int \frac{dx}{x^2 - 16} = \int \frac{dx}{x^2 - 4^2}$ $= \frac{1}{8} \log \left| \frac{x - 4}{x + 4} \right| + C$ |
