Inequality Integration

$\large \text{Since}, x^2 > \color{#3D99F6}{ x^8} > 0 \ \text{on} \ (0,1)$
$\large \Rightarrow x^2 + 1 > \color{#3D99F6}{x^8 +1 }> 0 + 1$
$\large \Rightarrow \dfrac{1}{1 + x^2} <\color{#3D99F6}{ \dfrac{1}{1 + x^8}} < 1$
$\large \Rightarrow \displaystyle\int_0^1 \dfrac{1}{1 + x^2}\text{dx} = \color{#D61F06}{ \dfrac{\pi}{4}} < \color{#3D99F6}{\displaystyle\int_0^1 \dfrac{1}{1 + x^8}\text{dx}} < \color{#D61F06}{ 1} = \displaystyle\int_0^1 1 \cdot \text{dx}$

n=1: $\ln 2 \approx 0.6931$

n=2: $\frac{\pi}{4} \approx 0.7854$

Example: $0.1^8 = 0.00000001$ which is much less than 0.1;

An effect of a tendency of zeroing $x^n$ when n increased in a limit for $\int_0^1 d x$ = 1 - 0 = 1.

In fact, $L = \displaystyle \int_0^1 \dfrac{1}{1 + x^8} \, dx$ $\approx$ 0.92465

Answer: $\boxed{\frac{\pi}{4} < L < 1}$