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

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

`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	0.5	0.3333	0.25	0.2	0.1667	0.1429

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 + 5xx0.5 + 0.3333 + 6xx0.25 + 0.2 + 5xx0.1667 + 0.1429)]`

`int y dx=(3xx1)/10 [6.5095]`

`int y dx=1.9529`

Solution by Weddle's Rule is `1.9529`

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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*0.5=2.5`

`y_2=0.3333`

`6y_3=6*0.25=1.5`

`y_4=0.2`

`5y_5=5*0.1667=0.8333`

`y_6=0.1429`

`int y dx=(3xx1)/10*[(1+2.5+0.3333+1.5+0.20.8333+0.1429)]`

`int y dx=(3xx1)/10*(6.5095)`

`int y dx=1.9529`

Solution by Weddle's Rule is `1.9529`

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