Question

$\underset{0}{\overset{3}{\int }}\frac{1}{x^2+9}dx.$

Answer 1

$$\begin{gathered} {\int }_0^3\frac{1}{x^2+9}dx \\ ={\int }_0^3\frac{1}{x^2+3^2}dx \\ =\frac{1}{3}{\left[{\tan }^{-1}\frac{x}{3}\right]}_0^3 \\ =\frac{1}{3}\left({\tan }^{-1}1-{\tan }^{-1}0\right) \\ =\frac{1}{3}\left(\frac{\mathrm{\pi }}{4}-0\right) \\ =\frac{\mathrm{\pi }}{12} \end{gathered}$$