Answer
$\int_0^{0.5} \dfrac{1}{\sqrt{1+x^4}} dx \approx 0.4969564$ (Error of about $10^{-5})$
Work Step by Step
Integrate the integral with respect to $ x $ as follows:
$\int_0^{0.5} \dfrac{1}{\sqrt{1+x^4}} dx=\int_0^{0.5} [1-\dfrac{x^4}{2}+\dfrac{3x^{8}}{8}-...] dx \\=0.5+\dfrac{(0.5)^5}{10}+\dfrac{3(0.5)^9}{24}-.... $
$\dfrac{5(0.5)^{13}}{16 \cdot 13} \approx 2.93 \times 10^{-6} $ ; this means that the proceeding term is greater than $10^{-5}$ and the first five terms of the series yields the accuracy.
Thus, $\int_0^{0.5} \dfrac{1}{\sqrt{1+x^4}} dx \approx 0.4969564$ (Error of about $10^{-5})$
