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!
 A: $\forall x \in [0,1], \frac{1}{1+x^8}<1$, 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
A: 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.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,}
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\begin{align}
&\color{#f00}{\int_{0}^{1}{\dd x \over 1 + x^{8}}}\ \stackrel{x^{8}\ \to\ x}{=}\
{1 \over 8}\int_{0}^{1}{x^{-7/8} \over 1 + x}\,\dd x =
{1 \over 8}\bracks{%
\int_{0}^{1}{\dd x \over 1 + x} - \int_{0}^{1}{1 - x^{-7/8} \over 1 + x}\,\dd x} \\[3mm] = &\
{1 \over 8}\,\ln\pars{2} -
{1 \over 8}\bracks{2\int_{0}^{1}{1 - x^{-7/8} \over 1 - x^{2}}\,\dd x -
\int_{0}^{1}{1 - x^{-7/8} \over 1 - x}\,\dd x}
\\[3mm] = &\
{1 \over 8}\,\ln\pars{2} -
{1 \over 8}\bracks{\int_{0}^{1}{x^{-1/2} - x^{-15/16} \over 1 - x}\,\dd x -
\int_{0}^{1}{1 - x^{-7/8} \over 1 - x}\,\dd x}
\\[3mm]  = &\
{1 \over 8}\,\ln\pars{2} -
{1 \over 8}\bracks{\int_{0}^{1}{1 - x^{-15/16} \over 1 - x}\,\dd x -
\int_{0}^{1}{1 - x^{-1/2} \over 1 - x}\,\dd x -
\int_{0}^{1}{1 - x^{-7/8} \over 1 - x}\,\dd x}
\\[3mm] = &\
{1 \over 8}\,\bracks{\ln\pars{2} - \Psi\pars{{1 \over 16}} + \Psi\pars{\half} +
\Psi\pars{{1 \over 8}} + \gamma}
\\[3mm] = &\
\color{#f00}{{1 \over 8}\bracks{\Psi\pars{{1 \over 8}} - \Psi\pars{{1 \over 16}} - \ln\pars{2}}} \approx 0.9247
\end{align}
where $\Psi$ is the Digamma function and $\gamma$ is the Euler-Mascheroni constant and with the use of the well known identities
$$
\left\lbrace\begin{array}{rcl}
\ds{\Psi\pars{z} + \gamma} & \ds{=} &
\ds{\int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t\,,\qquad\Re\pars{z} > 0}
\\[1mm]
\ds{\Psi\pars{\half}} & \ds{=} & \ds{-2\ln\pars{2} - \gamma}
\end{array}\right.
$$
A further reduction can be found with the identity ( see Gradshteyn &
Ryzhik table ):
\begin{align}
\Psi\pars{{p \over q}} & = - \gamma - \ln\pars{2q} -
{\pi \over 2}\,\cot\pars{\pi\,{p \over q}} +
2\sum_{k = 1}^{\left\lfloor\pars{q + 1}/2\right\rfloor - 1}
\cos\pars{2\pi k\,{p \over q}}\ln\pars{\sin\pars{{k\pi \over q}}}
\\[3mm] &\
q = 2,3,\ldots\,;\qquad p = 1,2,\ldots,\pars{q - 1}
\end{align}
A: The best and fastest way is to express the integrand as the sum of an infinite geometric series where $r=-x^8$, which converges in domain of the integral.  Since the question asks to approximate $L$, use the first couple terms of the series: \begin{align*}
L& \approx \int_0^1 1-x^8 \; \mathrm{d}x \\
&= x-\frac{x^9}{9} \bigg \rvert_0^1 \\
&=\frac{8}{9} \approx 0.9 \\
\end{align*}
The answer choices $L<1$ and $L>\frac{\pi}{4}$ are both correct.
