How to find $L = \int_0^1 \frac{dx}{1+{x^8}}$ Let $L = \displaystyle \int_0^1 \frac{dx}{1+{x^8}}$ . Then


*

*$L < 1$

*$L > 1$

*$L < \frac{\pi}{4}$

*$L > \frac{\pi}{4}$
I got some idea from this video link. But got stuck while evaluating the second integral. 
Please help!!
Thanks in advance!

$\forall x \in [0,1], \frac{1}{1+x^8}<1$, 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
