We'll calculate first, I0.
I0
= Int x^0dx/(x+1)
I0 = Int
dx/(x+1)
I0 = ln(x+1)
We'll
apply Leibniz-Newton to evaluate the value of definite integral
I0.
I0 = F(1) - F(0)
I0 =
ln(1+1) - ln(0+1)
I0 = ln2 - ln1, but ln 1 =
0
I0 = ln2
We'll determine
I1:
I1 = Int x^1dx/(x+1)
I1 =
Int xdx/(x+1)
I1 = Int dx - Int
dx/(x+1)
But Int dx/(x+1) = I0 = ln
2
I1 = x - I0
I1 = 1 - 0 -
I0
I1 = 1 - ln 2
We notice
that In+1 = Int x^ndx - In
In+1 + In = Int
x^ndx
In+1 + In = x^(n+1)/(n+1), for x = 0 to x =
1
The relation between the consecutive terms
is In+1 + In = 1/(n+1).
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