By Simpson rule taking $ n = 4 $ , the value of the integral $ \int^\limits{1}_{0} \frac{1}{1+x^{2}}dx $ is equal toLet $ y=\frac{\log x}{x} $ On differentiating w.r.t. $ x, $ we get $ \frac{dy}{dx}=\frac{x.\frac{1}{x}-\log x}{{{x}^{2}}} $ $ \Rightarrow $ $ \frac{dy}{dx}=\frac{1-\log x}{{{x}^{2}}} $ Put $ \frac{dy}{dx}=0 $ for maxima or minima $ \therefore $ $ x=e $ Now, $ \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{{{x}^{2}}\left( -\frac{1}{x} \right)-(1-\log x)2x}{{{({{x}^{2}})}^{2}}} $ $ =\frac{-3+2\log x}{{{x}^{3}}} $ and $ \therefore $ Function is maximum at $ x=e $ . $ \therefore $ $ {{y}_{\max }}=\frac{\log e}{e}=\frac{1}{e} $