Answer
$\approx 0.4969564$ with an error of about $10^{-5}$.
Work Step by Step
We integrate the integral with respect to $ x $ and take a lower limit as $0$ and upper limit as $0.4$.
$\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 $
or, $=0.5+\dfrac{(0.5)^5}{10}+\dfrac{3(0.5)^9}{24}-....$
But $\dfrac{5(0.5)^{13}}{16 \cdot 13} \approx 2.93 \times 10^{-6} $ and the proceeding term is greater than $10^{-5}$.
So, the first five terms of the series would give the accuracy.
$\int_0^{0.5} \dfrac{1}{\sqrt{1+x^4}} dx \approx 0.4969564$ with an error of about $10^{-5}$.
