Question
Evaluate $\int\frac{1}{\text{x}^2+16}\text{ dx}$

Answer

Since, $\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}$
Thus, $\int\frac{1}{\text{x}^2+16}\text{ dx}=\frac{1}{4}\tan^{-1}\Big(\frac{\text{x}}{4}\Big)+\text{C}$

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