Prove that I(n+2)+2010^2*I(n)=1/(n+1) if I(n) is the definite integral of y=x^n/(x^2+2010^2) for x=0 to x=1 .

We'll write the definite integral of
I(n):

I(n) = Int x^n
dx/(x^2+2010^2)

We'll multiply I(n) by
2010^2:

2010^2I(n) = Int 2010^2*x^n dx/(x^2+2010^2)
(1)

We'll write the definite integral of
I(n+2):

I(n+2) = Int x^(n+2) dx/(x^2+2010^2)
(2)

We'll apply the property of integral to be
additive:

(1) + (2) = Int [x^(n+2) +
2010^2*x^n]dx/(x^2+2010^2)

Well factorize by
x^n:

Int [x^n(x^2 +
2010^2)]dx/(x^2+2010^2)

We'll simplify by (x^2+2010^2) and
we'll get:

Int x^n dx =
x^(n+1)/(n+1)

For x = 0 => 0^(n+1)/(n+1) =
0

For x = 1 =>
1/(n+1)

 We'll apply
Leibniz-Newton:

Int x^n dx = F(1) -
F(0)

Int x^n dx =
1/(n+1)

I(n+2) + 2010^2I(n) = Int x^n dx
= 1/(n+1) q.e.d.