Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 286: 32

$\frac{1}{2}tan^{-1}(\frac{2}{\sqrt{3}})$

$\int^{\frac{1}{\sqrt{3}}}_{0}\frac{1}{1-4x^2}dx$ =$\int^{\frac{1}{\sqrt{3}}}_{0}\frac{1}{1-2^2x^2}dx$ =$\frac{1}{2}tan^{-1}(\frac{2}{\sqrt{3}})-\frac{1}{2}tan^{-1}(0)$ =$\frac{1}{2}tan^{-1}(\frac{2}{\sqrt{3}})$