Estimation of $\int_0^1 \frac{1}{1+x^4}dx$

Question

Prove that $\dfrac34<\displaystyle\int_0^1 \frac{1}{1+x^4}\,\mathrm dx<\dfrac {9}{10}$.

My working:

We can easily prove that $$\begin{align} \frac{1}{1+x^2}&<\frac{1}{1+x^4}<1-x^4+x^8,\forall x\in (0,1) \\ \implies\int_0^1\frac{1}{1+x^2}\,\mathrm dx&<\int_0^1\frac{1}{1+x^4}\,\mathrm dx<\int_0^1(1-x^4+x^8)\,\mathrm dx \\ \implies\frac34&<\frac{\pi}{4}<\int_0^1\frac{1}{1+x^4}\,\mathrm dx<(1-\frac 15+\frac 19)=\frac{41}{45} \end{align}$$

But unfortunately $\dfrac {9}{10}<\dfrac{41}{45}$.

Answer 1

We can easily prove that $$\begin{align} 1-x^3&<\frac{1}{1+x^4}<1-\frac{x^4}{2},\forall x\in (0,1) \\ \implies\int_0^1 1-x^3\,\mathrm dx&<\int_0^1\frac{1}{1+x^4}\,\mathrm dx<\int_0^11-\frac{x^4}{2}\,\mathrm dx \\ \implies\frac34&<\int_0^1\frac{1}{1+x^4}\,\mathrm dx<1-\frac {1}{10}=\frac{9}{10} \end{align}$$