1) Trapezoidal method 2) Simpsons 1/3 rd method 3) Simpsons 3/8 th method

Compare result with exact values

Let $a = -1, b = 1$ and $n = 6$

$Let \space y=\dfrac 1{1+x^2}\\ \therefore h=\dfrac {b-a}n=\dfrac {1-(-1)}6=\dfrac 13$

1) Trapezoidal rule

$\int\limits_a^bf(x)dx=\dfrac h2[(y_0+y_6)+2(y_1+y_2+y_3+y_4+y_5)]\\ \therefore \int\limits^1_{-110}\dfrac {dx}{1+x^2}=\dfrac {\frac 13}2\Bigg[\Bigg(\dfrac 12+\dfrac 12\Bigg)+2\Bigg(\dfrac 9{13}+\dfrac 9{10}+1+\dfrac 9{10}+\dfrac 9{13}\Bigg)\Bigg]\\ =\dfrac 16\Bigg[1+2\times \dfrac {272}{65}\Bigg] \\ =1.5615$

2) Simpsons 1/3rd rule

$\int\limits_a^bf(x)dx=\dfrac h3[(y_0+y_6)+4(y_1+y_2+y_5)+2(y_3+y_4)] \\ =\dfrac {\frac 13}2\Bigg[\Bigg(\dfrac 12+\dfrac 12\Bigg)+4\Bigg(\dfrac 9{13}+1+\dfrac 9{13}\Bigg)+2\Bigg(9{10}+\dfrac 9{10}\Bigg)\Bigg]\\ =\dfrac 19\Bigg[1+4\times \dfrac {31}{13}+\dfrac {2\times9}5\Bigg]\\ =1.5709$

3) Simpsons 3/8th rule

$\int\limits_a^bf(x)dx=\dfrac {3h}8[(y_0+y_6)+3(y_1+y_2+y_4+y_5)+2(y_3)] \\ =\dfrac {3\times \frac 13}8\Bigg[\Bigg(\dfrac 12+\dfrac 12\Bigg)+3\Bigg(\dfrac 9{13} +\dfrac 9{10} + \dfrac 9{10} + \dfrac 9{13}\Bigg)+2(1)\Bigg]\\ =\dfrac 18\Bigg[1+3\times \dfrac {207}{65}+2 \Bigg] =1.5692$

Exact solution

$\int\limits^1_{-1}\dfrac 1{1+x^2}dx=[\tan^{-1}x]^1_{-1}\\ =\tan^{-1}1-\tan^{-1}(-1)\\ =\dfrac \pi4+\dfrac \pi4=\dfrac \pi2\\ =1.57079$