Find the approximated integral value of an equation 1/(x+1)^2 using Weddle's rule
a = 0 and b = 6
Interval n = 6

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

`a=0`

`b=6`

`Delta x =(b-a)/n=(6 - 0)/6=1`

The value of table for `x` and `f(x)`

`x`	`f(x)`
`x_0=0`	`f(x_(0))=f(0)=1`
`x_1=1`	`f(x_(1))=f(1)=0.25`
`x_2=2`	`f(x_(2))=f(2)=0.1111`
`x_3=3`	`f(x_(3))=f(3)=0.0625`
`x_4=4`	`f(x_(4))=f(4)=0.04`
`x_5=5`	`f(x_(5))=f(5)=0.0278`
`x_6=6`	`f(x_(6))=f(6)=0.0204`

Method-1:
Using Weddle's Rule


`int f(x) dx=(3Delta x )/10 [(f(x_(0))+5f(x_(1))+f(x_(2))+6f(x_(3))+f(x_(4))+5f(x_(5))+f(x_(6)))]`

`f(x_(0))=1`

`5f(x_(1))=5*0.25=1.25`

`f(x_(2))=0.1111`

`6f(x_(3))=6*0.0625=0.375`

`f(x_(4))=0.04`

`5f(x_(5))=5*0.0278=0.1389`

`f(x_(6))=0.0204`

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

`=(3xx1)/10*(2.9354)`

`=0.8806`

Solution by Weddle's Rule is `0.8806`

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Method-2:
Using Weddle's Rule


`int f(x) dx=(3Delta x )/10 [(f(x_(0))+5f(x_(1))+f(x_(2))+6f(x_(3))+f(x_(4))+5f(x_(5))+f(x_(6)))]`

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

`=(3xx1)/10 [2.9354]`

`=0.8806`

Solution by Weddle's Rule is `0.8806`

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