Answer
\[\frac{π}{12}\]
Work Step by Step
Let \[I=\int_{0}^{\frac{\sqrt 3}{4}}\frac{dx}{1+16x^2}\]
\[I=\frac{1}{16}\int_{0}^{\frac{\sqrt 3}{4}}\frac{dx}{x^2+\left(\frac{1}{4}\right)^2}\;\;\;...(1)\]
We will use the formula \[\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}\;\;\;...(2)\]
Using (2) in (1)
\[I=\frac{1}{16}\left[4\tan^{-1}(4x)\right]_{0}^{\frac{\sqrt 3}{4}}\]
\[I=\frac{1}{4}\left[\tan^{-1}(\sqrt 3)-\tan^{-1} (0)\right]\]
\[I=\frac{1}{4}\left[\frac{π}{3}\right]\]
Hence \[I=\frac{π}{12}\]
