2. Find Solution of an equation 1/(x+1) using Simpson's 3/8 rule
x1 = 0 and x2 = 1
Interval N = 5

Solution:
Equation is `f(x)=(1)/(x+1)`.

`h = (b-a)/N`

`h = (1 - 0) / 5 = 0.2`

The value of table for `x` and `y`

x	0	0.2	0.4	0.6	0.8	1
y	1	0.8333	0.7143	0.625	0.5556	0.5

Method-1:
Using Simpson's `3/8` Rule

`int y dx=(3h)/8 [(y_0+y_5)+2(y_3)+3(y_1+y_2+y_4)]`

`int y dx=(3xx0.2)/8 [(1 +0.5)+2xx(0.625)+3xx(0.8333+0.7143+0.5556)]`

`int y dx=(3xx0.2)/8 [(1 +0.5)+2xx(0.625)+3xx(2.1032)]`

`int y dx=(3xx0.2)/8 [(1.5)+(1.25)+(6.3095)]`

`int y dx=0.6795`

Solution by Simpson's `3/8` Rule is `0.6795`

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Method-2:
Using Simpson's `3/8` Rule

`int y dx=(3h)/8 [y_0+3y_1+3y_2+2y_3+3y_4+y_5]`

`y_0=1`

`3y_1=3*0.8333=2.5`

`3y_2=3*0.7143=2.1429`

`2y_3=2*0.625=1.25`

`3y_4=3*0.5556=1.6667`

`y_5=0.5`

`int y dx=(3xx0.2)/8 *(1+2.5+2.1429+1.25+1.6667+0.5)`

`int y dx=(3xx0.2)/8 *(9.0595)`

`int y dx=0.6795`

Solution by Simpson's `3/8` Rule is `0.6795`

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