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!

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.
