You should notice that if `x in (0,1)` the following inequality holds such that:

`0 < (x^(2n))/(1 + x) < (x^(2n))/1`

You need to integrate such that:

`int_0^1 0 dx < int_0^1 (x^(2n))/(1 + x) dx < int_0^1 (x^(2n))/1 dx`

`0 < int_0^1 (x^(2n))/(1 + x) dx < (x^(2n+1))/(2n+1)|_0^1`

Using the fundamental theorem of calculus yields:

`0 < int_0^1 (x^(2n))/(1 + x) dx < (1^(2n+1) - 0^(2n+1))/(2n+1)`

`0 < int_0^1 (x^(2n))/(1 + x) dx < (1-0)/(2n+1)`

`0 < int_0^1 (x^(2n))/(1 + x) dx < 1/(2n+1)`

**Hence, the inequality may be proved without evaluating the integral, thus `0 < int_0^1f_n(x) dx< 1/(2n+1).` **