How to find $L = \int_0^1 \frac{dx}{1+{x^8}}$
$\forall x \in [0,1], \frac{1}{1+x^8}<1$, so $L<1$.
$\forall x \in [0,1], \frac{1}{1+x^8}>\frac{1}{1+x^2}$, so $L>\int_0^1 \frac{dx}{1+{x^2}}=\frac{\pi}{4}$
How to evaluate this integral :
$$\begin{align} \int \frac{1}{1+x^8}dx &= \int (1+x^8)^{-1}dx \\&= \int_0^x(1+t^8)^{-1}dt +a \\&= \int_0^{x^8} (1+t)^{-1}d(t^{1/8})+b \\ &=\frac{1}{8}\int_0^{x^8}(1+t)^{-1}t^{1/8 -1}dt+c \\ &= \frac{1}{8}\int_0^{1} (1+x^8t)^{-1} (x^8t)^{1/8-1} d(x^8t)+d \tag{1} \\ &= \frac{x}{8} \int_0^1(1+x^8t)^{-1}x^{1-8}t^{1/8-1}x^8dt+e \tag{2} \\ &= x\,{}_2F_1 \left( 1,\frac{1}{8},\frac{9}{8},-x^8 \right) +f\end{align}$$
So $L={}_2F_1 \left( 1,\frac{1}{8},\frac{9}{8},-1 \right)$.
$(1)$ I substitute $t$ with $x^8t$
$(2)$ Here $x$ is just a constant
You can get to a closed form by partial fractions without too much trouble, it just takes adherence to a protocol that keeps things relatively simple. First, expand into linear complex fractions $$\frac1{x^8+1}=\frac1{\prod_{k=0}^7(x-\omega_k)}=\sum_{k=0}^7\frac{A_k}{x-\omega_k}$$ Where $\omega_k=e^{i\theta_k}=\cos\theta_k+i\sin\theta_k$ and $\theta_k=\frac{\pi(2k+1)}8$. Then $$\lim_{x\rightarrow\omega_k}\frac{x-\omega_k}{x^8+1}=\lim_{x\rightarrow\omega_k}\frac1{8x^7}=\frac1{8\omega_k^7}=-\frac{\omega_k}8=\lim_{x\rightarrow\omega_k}\sum_{j=0}^7A_j\frac{x-\omega_k}{x-\omega_j}=\sum_{j=0}^7A_j\delta_{jk}=A_k$$ Just as @Robert Israel said. But then we diverge for a while. Here I would use the symmetry $$\theta_{7-k}=\frac{\pi(15-2k)}8=2\pi-\theta_k$$ So $\omega_{7-k}=\omega_k^*$. Then $$\begin{align}\frac1{x^8+1}&=-\frac18\sum_{k=0}^7\frac{\omega_k}{x-\omega_k}=-\frac18\sum_{k=0}^3\left(\frac{\omega_k}{x-\omega_k}+\frac{\omega_k^*}{x-\omega_k^*}\right)\\ &=-\frac18\sum_{k=0}^3\frac{2x\cos\theta_k-2}{x^2-2x\cos\theta_k+1}=-\frac14\sum_{k=0}^3\frac{(x-\cos\theta_k)\cos\theta_k-\sin^2\theta_k}{(x-\cos\theta_k)^2+\sin^2\theta_k}\end{align}$$ And now we can integrate more smoothly $$\begin{align}\int\frac{dx}{x^8+1}&=-\frac14\sum_{k=0}^3\int\frac{(x-\cos\theta_k)\cos\theta_k-\sin^2\theta_k}{(x-\cos\theta_k)^2+\sin^2\theta_k}\\ &=-\frac14\sum_{k=0}^3\left[\frac12\cos\theta_k\ln\left(x^2-2x\cos\theta_k+1\right)-\sin\theta_k\tan^{-1}\left(\frac{x-\cos\theta_k}{\sin\theta_k}\right)\right]+C\end{align}$$ Now we can use another symmetry $$\theta_{3-k}=\frac{\pi(7-2k)}8=\pi-\theta_k$$ So $\cos\theta_{3-k}=-\cos\theta_k$ and $\sin\theta_{3-k}=\sin\theta_k$. We can add the arctangents because $$\begin{align}\tan^{-1}\left(\frac{x-\cos\theta_k}{\sin\theta_k}\right)+\tan^{-1}\left(\frac{x+\cos\theta_k}{\sin\theta_k}\right)&=\tan^{-1}\left(\frac{\frac{x-\cos\theta_k}{\sin\theta_k}+\frac{x+\cos\theta_k}{\sin\theta_k}}{1-\frac{x^2-\cos^2\theta_k}{\sin^2\theta_k}}\right)\\ &=\tan^{-1}\left(\frac{2x\sin\theta_k}{1-x^2}\right)\end{align}$$ And subtracting logarithms leads to ratios, so $$\int_0^1\frac{dx}{x^8+1}=-\frac14\sum_{k=0}^1\left[\frac12\cos\theta_k\ln\left(\frac{x^2-2x\cos\theta_k+1}{x^2-2x\cos\theta_k+1}\right)-\sin\theta_k\tan^{-1}\left(\frac{2x\sin\theta_k}{1-x^2}\right)\right]_0^1$$ We can now use $$\cos\theta_0=\sin\theta_1=\frac{\sqrt{2+\sqrt2}}2$$ $$\cos\theta_1=\sin\theta_0=\frac{\sqrt{2-\sqrt2}}2$$ $$\lim_{x\rightarrow\infty}\tan^{-1}x=\frac{\pi}2$$ And $\tan0=\ln1=0$ to get to $$\begin{align}\int_0^1\frac{dx}{x^8+1}&=-\frac{\sqrt{2+\sqrt2}}{16}\ln\left(\frac{2-\sqrt{2+\sqrt2}}{2+\sqrt{2+\sqrt2}}\right)+\frac{\sqrt{2-\sqrt2}\cdot\pi}{16}\\ &-\frac{\sqrt{2-\sqrt2}}{16}\ln\left(\frac{2-\sqrt{2-\sqrt2}}{2+\sqrt{2-\sqrt2}}\right)+\frac{\sqrt{2+\sqrt2}\cdot\pi}{16}\end{align}$$ After checking that it matches the numerical result of $0.924651705775538$ we are satisfied with the result, same as @Robert Israel, but we have shown this time how the symmetries of the problem keep the expansions from getting out of hand. As has already been pointed out, this is between the easy bounds of $\frac{\pi}4$ and $1$.
EDIT: I just had to try and get the closed form to fit on a single line. $$\int_0^1\frac{dx}{x^8+1}=\frac{\sqrt{2+\sqrt2}}{16}\left(\sqrt8\sinh^{-1}\sqrt[4]2+\pi\right)+\frac{\sqrt{2-\sqrt2}}{16}\left(\sqrt8\cosh^{-1}\sqrt[4]2+\pi\right)$$ Or even $$\int_0^1\frac{dx}{x^8+1}=\frac{\sqrt{2+\sqrt2}}{8\sqrt2}\left(2\sinh^{-1}\sqrt[4]2+\pi\right)+\frac{\sqrt{2-\sqrt2}}{4\sqrt2}\cosh^{-1}\sqrt[4]2$$
There is a "closed-form" expression for this integral. Using partial fractions, $$ \dfrac{1}{1+x^8} = - \dfrac{1}{8} \sum_{\omega} \dfrac{\omega}{x - \omega}$$ where the sum is over the roots of the polynomial $x^8 + 1$. Then integrate: $$ \int_0^1 \dfrac{dx}{1+x^8} = -\dfrac{1}{8} \sum_\omega \omega (\log(1-\omega) - \log(-\omega))$$ where we use the principal branch of the logarithm (note that the line segments from $-\omega$ to $1-\omega$ never cross the negative real axis, which is the branch cut of this function, so this works). These roots can be expressed explicitly as $\left(\pm \sqrt {2+\sqrt {2}}\pm i\sqrt {2-\sqrt {2}}\right)/2$ and $\left(\pm \sqrt {2-\sqrt {2}}\pm i\sqrt {2+\sqrt {2}}\right)/2$. The final result is a rather messy, but explicit, expression in terms of $16$ complex logarithms. It can then be simplified to
$$\frac{1}{16} \left( \ln \left( \left( 4-2\,\sqrt {2} \right) \sqrt {2- \sqrt {2}}-4\,\sqrt {2}+7 \right) +\pi \right) \sqrt {2-\sqrt {2}}+\frac{1}{16} \left( \ln \left( \left( 4+2\,\sqrt {2} \right) \sqrt {2+\sqrt {2}}+4\,\sqrt {2}+7 \right) +\pi \right) \sqrt {2+\sqrt {2}}$$
The numerical value turns out to be approximately $.9246517057$.
Of course, this is not the best way to answer the original question.