Answer
\[\frac{\pi }{{12}}\]
Work Step by Step
\[\begin{gathered}
\int_1^{\sqrt 3 } {\frac{{dx}}{{1 + {x^2}}}} \hfill \\
\hfill \\
{\text{integrating}} \hfill \\
\hfill \\
\int_1^{\sqrt 3 } {\frac{{dx}}{{1 + {x^2}}}} = \left[ {{{\tan }^{ - 1}}x} \right]_1^{\sqrt 3 } \hfill \\
\hfill \\
{\text{Use the fundamental theorem of calculus}} \hfill \\
\hfill \\
{\tan ^{ - 1}}\left( {\sqrt 3 } \right) - {\tan ^{ - 1}}\left( 1 \right) \hfill \\
\hfill \\
{\text{Simplify}} \hfill \\
\hfill \\
\frac{\pi }{3} - \frac{\pi }{4} \hfill \\
\hfill \\
\frac{\pi }{{12}} \hfill \\
\end{gathered} \]
