Prove In+2-In=(3^(n+1)-2^(n+1))/(n+1) if In=definite integral of y= x^n/(x^2-1) when x=2 to x=3.

We'll re-write the difference  In+2 - In, based on the
identity given by enunciation: In = Int x^ndx/(x^2 -
1)

In+2 - In = Int x^(n+2)dx/(x^2 - 1) - Int x^ndx/(x^2 -
1)

We'll re-write the power x^(n+2) =
x^n*x^2

In+2 - In = Int x^n*x^2dx/(x^2 - 1) - Int
x^ndx/(x^2 - 1)

We'll use the property of addition of
integrals:

In+2 - In = Int (x^n*x^2 - x^n)dx/(x^2 -
1)

We'll factorize the numerator by
x^n:

In+2 - In = Int x^n*(x^2 - 1)dx/(x^2 -
1)

We'll simplify:

In+2 - In =
Int x^n dx

We'll apply Leibniz Newton formula to evaluate
the definite integral:

Int x^n dx = F(3) -
F(2)

F(3) = 3^(n+1)/(n+1)

F(2)
= 2^(n+1)/(n+1)

F(3) - F(2) = 3^(n+1)/(n+1) -
2^(n+1)/(n+1)

F(3) - F(2) = [3^(n+1) -
2^(n+1)]/(n+1)

The identity In+2 - In =
[3^(n+1) - 2^(n+1)]/(n+1) is verified, if In = Int x^ndx/(x^2 - 1), when the limits of
integration are x = 2 and x = 3.