Question

Evaluate: \[\int\frac{1}{x^2 + 16}\text{ dx }\]

Answer 1

\[\int\frac{1}{x^2 + 16}dx = \int\frac{1}{x^2 + 4^2}dx\]
 \[ = \frac{1}{4} \tan^{- 1} \frac{x}{4} + c\]
\[\text{ Hence,} \int\frac{1}{x^2 + 16}dx = \frac{1}{4} \tan^{- 1} \frac{x}{4} + c\]