Answer
$$1.109.$$
Work Step by Step
Given$$\int_{0}^{3} \frac{d x}{x^{4}+1}, \quad N=6 $$
Since $\Delta x=\dfrac{b-a}{N}=0.5$ , then by using Simpson’s rule
\begin{align*}
S_{n}&=\dfrac{\Delta x}{3}\left[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3).....+4f(x_{n-1})+f(x_n)\right]\\
S_{4}&=\dfrac{\Delta x}{3}\left[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+4f(x_5) +f(x_{6}) \right] \\
&=\dfrac{1}{6}\left[f(0)+4f(0.5)+2f(1.5)+4f(2)+2f(2.5) +f(3) \right]\\
&\approx 1.109.
\end{align*}
