# Verify if I(n+4)+I(n)=1/(n+1), if I(n)=Integral of x^n/(x^4+1)?limits of integration ar x=0 and x=1

justaguide | Certified Educator

It is given that I(n) = Int[x^n/(x^4+1) dx], x = 0 to x = 1

I(n+4) + I(n)

=> Int[x^(n+4)/(x^4+1) dx], x = 0 to x = 1 + Int[x^n/(x^4+1) dx], x = 0 to x = 1

=> Int[x^(n+4)/(x^4+1) dx] + Int[x^n/(x^4+1) dx], x = 0 to x = 1

=> Int[(x^(n+4)/(x^4+1) + x^n/(x^4+1)) dx], x = 0 to x = 1

=> Int[(x^(n+4) + x^n)/(x^4+1) dx], x = 0 to x = 1

=> Int[(x^n(x^4 + 1)/(x^4+1) dx], x = 0 to x = 1

=> Int[x^n dx], x = 0 to x = 1

=> [x^(n + 1)]/(n + 1), x = 0 to x = 1

=> 1^(n + 1)/(n + 1) - 0^(n + 1)/(n + 1)

=> 1/(n+1)

This proves I(n+4)+I(n)=1/(n+1)

giorgiana1976 | Student

We'll write I(n+4)+I(n) using the property of integral to be additive:

I(n+4)+I(n) = Int x^(n+4)dx/(x^4 + 1) + Int x^n dx/(x^4+1)

I(n+4)+I(n) = Int (x^n*x^4 + x^n)dx/(x^4+1)

We'll factorize by x^n at numerator:

I(n+4)+I(n) = Int x^n*(x^4 + 1)dx/(x^4+1)

We'll simplify and we'll get:

I(n+4)+I(n) = Int x^ndx

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

Int x^ndx = F(1) - F(0)

Int x^ndx = 1^(n+1)/(n+1) - 0^(n+1)/(n+1)

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

We notice that the identity I(n+4)+I(n) = 1/(n+1) is verified.