Given I = Integral x^n/(x^n+1) demonstrate that I

We'll write the definite integral of the
function:

Int x^ndx/(x^n+1)(x=0 to x=2) = Intx^ndx/(x^n+1)(x=0 to
x=1) + Int x^ndx/(x^n+1)(x=1 to x=2)

Since n is natural,
then:

x^n < x^n + 1

x^n/(x^n +
1) < 1

We'll integrate both
sides:

Int x^ndx/(x^n + 1) < Int
dx

Also, Int x^ndx/(x^n + 1) < Int x^n
dx

Intx^ndx/(x^n+1)(x=0 to x=1) + Int x^ndx/(x^n+1)(x=1 to x=2)
< Int x^n dx(x=0 to x=1) + Int dx(x=1 to x=2)

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

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

Int dx(x=1 to x=2) = x = 2 - 1 =
1

Therefore I = Int x^ndx/(x^n + 1) < 1/(n+1) +
1.