Let $J={\int }_0^1\frac{x}{1+x^8}dx$ Consider the following assertions:

$\mathrm{I}$. $J>\frac{1}{4}$

$\mathrm{II}$. $J<\frac{\pi }{8}$

Then,

As we know, $1+x^8<2,\forall x\in \left(0,1\right)$

$\Rightarrow \frac{1}{2}<\frac{1}{1+x^8},\forall x\in \left(0,1\right)$

$\Rightarrow \frac{x}{2}<\frac{x}{1+x^8},\forall x\in \left(0,1\right)$

$\Rightarrow {\int }_0^1\frac{x}{2}dx<{\int }_0^1\frac{x}{1+x^8}dx$

$\Rightarrow {\int }_0^1\frac{x}{1+x^8}dx>\frac{1}{4}$

$\Rightarrow J>\frac{1}{4}$

Similarly, as we know, $1+x^8<1+x^4,\forall x\in \left(0,1\right)$

$\Rightarrow {\int }_0^1\frac{x}{1+x^4}dx<{\int }_0^1\frac{x}{1+x^8}dx$

$\Rightarrow J>\frac{1}{2}{\left[{\tan }^{-1}\left(x^2\right)\right]}_0^1$

$\Rightarrow J>\frac{\pi }{8}$

Statement $J={\int }_0^1\frac{x}{1+x^8}dx>\frac{1}{4}$ is true, while the statement $J={\int }_0^1\frac{x}{1+x^8}dx<\frac{\pi }{8}$ is not true.