3. Find Solution of an equation 1/(x+1)^2 using Weddle's rule
x1 = 0 and x2 = 6
Interval N = 6

Solution:
Equation is `f(x)=(1)/(x+1)^2`.

`h = (b-a)/N`

`h = (6 - 0) / 6 = 1`

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

x	0	1	2	3	4	5	6
y	1	0.25	0.1111	0.0625	0.04	0.0278	0.0204

Method-1:
Using Weddle's Rule
`int y dx=(3h)/10 [(y_0+5y_1+y_2+6y_3+y_4+5y_5+y_6)]`

`int y dx=(3xx1)/10 [(1 + 5xx0.25 + 0.1111 + 6xx0.0625 + 0.04 + 5xx0.0278 + 0.0204)]`

`int y dx=(3xx1)/10 [2.9354]`

`int y dx=0.8806`

Solution by Weddle's Rule is `0.8806`

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Method-2:
Using Weddle's Rule
`int y dx=(3h)/10 [(y_0+5y_1+y_2+6y_3+y_4+5y_5+y_6)]`

`y_0=1`

`5y_1=5*0.25=1.25`

`y_2=0.1111`

`6y_3=6*0.0625=0.375`

`y_4=0.04`

`5y_5=5*0.0278=0.1389`

`y_6=0.0204`

`int y dx=(3xx1)/10*[(1+1.25+0.1111+0.375+0.040.1389+0.0204)]`

`int y dx=(3xx1)/10*(2.9354)`

`int y dx=0.8806`

Solution by Weddle's Rule is `0.8806`

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