# 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.

### 1 Answer | Add Yours

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.**