Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 320: 32

$\approx0.4285$

$\int_{0}^{1/\sqrt 3} \frac{dx}{1+4x^2}=\frac{1}{2}\cdot\int_{0}^{1/\sqrt 3} \frac{2dx}{1+(2x)^2}=\frac{1}{2}\cdot[arctan(2x)]_{0}^{1/\sqrt 3}=\frac{1}{2}\cdot[arctan(\frac{2\cdot1}{\sqrt 3})]-\frac{1}{2}\cdot[arctan({2\cdot0})]\approx0.4285-0\approx0.4285$