# Given I = Integral x^n/(x^n+1) demonstrate that I <= 1+1/(n+1). The limits of integration are x=0 to x=2 and n is natural number.

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### 1 Answer

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.**