Compute: $\int_{0}^{1}\frac{x^4+1}{x^6+1} dx$ I'm trying to compute: $$\int_{0}^{1}\frac{x^4+1}{x^6+1}dx.$$
I tried to change $x^4$ into $t^2$ or $t$, but it didn't work for me.
Any suggestions?
Thanks!
 A: The denominator of the integrand $f(x):=\dfrac{x^{4}+1}{x^{6}+1}$ may be factored as
\begin{eqnarray*}
x^{6}+1 &=&\left( x^{2}+1\right) \left( x^{4}-x^{2}+1\right)  \
&=&\left( x^{2}+1\right) \left( x^{2}-\sqrt{3}x+1\right) \left( x^{2}+\sqrt{3
}x+1\right) 
\end{eqnarray*}
If you expand $f(x)$ you get
$$\begin{eqnarray*}
f(x) &=&\frac{2}{3}\frac{1}{x^{2}+1}+\frac{1}{6}\frac{1}{x^{2}-\sqrt{3}x+1}+
\frac{1}{6}\frac{1}{x^{2}+\sqrt{3}x+1} \\
&=&\frac{2}{3}\frac{1}{x^{2}+1}+\frac{2}{3}\frac{1}{\left( 2x-\sqrt{3}
\right) ^{2}+1}+\frac{2}{3}\frac{1}{\left( 2x+\sqrt{3}\right) ^{2}+1}.
\end{eqnarray*}$$
Since
$$
\int \frac{1}{x^{2}+1}dx=\arctan x
$$
and
$$
\begin{eqnarray*}
\int \frac{1}{\left( ax+b\right) ^{2}+1}dx &=&\int \frac{1}{a\left(
u^{2}+1\right) }\,du=\frac{1}{a}\arctan u \\
&=&\frac{1}{a}\arctan \left( ax+b\right), 
\end{eqnarray*}
$$
we have
$$
\begin{eqnarray*}
\int_{0}^{1}\frac{x^{4}+1}{x^{6}+1}dx &=&\frac{2}{3}\int_{0}^{1}\frac{1}{%
x^{2}+1}dx+\frac{2}{3}\int_{0}^{1}\frac{1}{\left( 2x-\sqrt{3}\right) ^{2}+1}%
dx \\
&&+\frac{2}{3}\int_{0}^{1}\frac{1}{\left( 2x+\sqrt{3}\right) ^{2}+1}dx \\
&=&\frac{2}{3}\arctan 1+\frac{2}{3}\left( \frac{1}{2}\arctan \left( 2-\sqrt{3%
}\right) -\frac{1}{2}\arctan \left( -\sqrt{3}\right) \right)  \\
&&+\frac{2}{3}\left( \frac{1}{2}\arctan \left( 2+\sqrt{3}\right) -\frac{1}{2}%
\arctan \left( \sqrt{3}\right) \right)  \\
&=&\frac{1}{6}\pi +\frac{1}{3}\left( \arctan \left( 2-\sqrt{3}\right)
+\arctan \left( \sqrt{3}\right) \right)  \\
&&+\frac{1}{3}\left( \arctan \left( 2+\sqrt{3}\right) -\arctan \left( \sqrt{3%
}\right) \right)  \\
&=&\frac{1}{6}\pi +\frac{1}{3}\left( \arctan \left( 2-\sqrt{3}\right)
+\arctan \left( 2+\sqrt{3}\right) \right)  \\
&=&\frac{1}{6}\pi +\frac{1}{6}\pi  \\
&=&\frac{1}{3}\pi, 
\end{eqnarray*}
$$
because$^1$
$$
\arctan \left( 2-\sqrt{3}\right) +\arctan \left( 2+\sqrt{3}\right) =\frac{1}{
2}\pi. 
$$

$^1$We apply the arctangent additional formula to $u=2-\sqrt{3}$ and $v=2+\sqrt{3}$
$$
\arctan u+\arctan v=\arctan \frac{u+v}{1-uv}.
$$
Since the product $uv=1$ and $\arctan \left( 2-\sqrt{3}\right) >0,\arctan
\left( 2+\sqrt{3}\right) >0$, we get on the right $\arctan \dfrac{4}{1-1}=
\dfrac{\pi }{2}.$
A: A not so simple but funny way to compute it :
Denote I the value we are looking for.
With a power series expansion of the integrand, we have
$$ I = 1 + 2\sum_{n=1}^\infty \frac{(-1)^n}{36n^2-1} $$
With another series expansion and interversion of the summation, we have
$$ I = 1 + 2\sum_{k=1}^\infty 6^{-2k}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^{2k}} $$
We recognize the Dirichlet Eta function evaluated at even integers, so
$$ I = 1+\sum_{k=1}^\infty \frac{(2^{2k}-2)\pi^{2k}}{6^{2k}}\frac{|B_{2k}|}{(2k)!} $$
Recognizing the well-known series exansion
$$ 1 - \frac x2 \mathrm{cot} \frac x2 = \sum_{k=1}^\infty \frac{|B_{2k}| x^{2k}}{(2k)!},$$
we have
$$ I = 1 + f(\pi / 3) - 2f(\pi/6), $$ where $f$ is the above function.
The trigonometric computation is not trivial but we eventually find
$$ I = \frac{\pi}{3} $$
A: First substitute $x=\tan\theta$. Simplify the integrand, noticing that $\sec^2\theta$ is a factor of the original denominator. Use the identity connecting $\tan^2\theta$ and $\cos2\theta$ to write the integrand in terms of $\cos^22\theta$. Now the substitution $t=\tan2\theta$ reduces the integral to a standard form, which proves $\pi/3$ to be the correct answer. This method seems rather roundabout in retrospect, but it requires only natural substitutions, standard trigonometric identities, and straightforward algebraic simplification. 
A: Edited Here is a much simpler version of the previous answer.
$$\int_0^1 \frac{x^4+1}{x^6+1}dx =\int_0^1 \frac{x^4-x^2+1}{x^6+1}dx+ \int_0^1 \frac{x^2}{x^6+1}dx$$
After canceling the first fraction, and subbing $y=x^3$ in the second we get:
$$\int_0^1 \frac{x^4+1}{x^6+1}dx =\int_0^1 \frac{1}{x^2+1}dx+ \frac{1}{3}\int_0^1 \frac{1}{y^2+1}dy = \frac{\pi}{4}+\frac{\pi}{12}=\frac{\pi}{3} \,.$$
P.S. Thanks to Zarrax for pointing the stupid mistakes I did...
A: one way is partial fractions on
$$
\frac{x^4+1}{x^6+1}=\frac{(x-e^{\pi i/4})(x-e^{3\pi i/4})(x-e^{5\pi i/4})(x-e^{7\pi i/4})}{(x-e^{\pi i/6})(x-e^{3\pi i/6})(x-e^{5\pi i/6})(x-e^{7\pi i/6})(x-e^{9\pi i/6})(x-e^{11\pi i/6})}
$$
$$
=\frac{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}{(x^2+1)(x^2+\sqrt{3}x+1)(x^2-\sqrt{3}x+1)}
$$
A: Note that
$$I=\int_{0}^{1}\frac{x^4+1}{x^6+1}\,dx\quad\stackrel{\large x\,\mapsto\,\frac{1}{x}}\Longrightarrow\quad I=\int_{1}^{\infty}\frac{x^4+1}{x^6+1}\,dx\quad\Longrightarrow\quad I=\frac{1}{2}\int_{0}^{\infty}\frac{x^4+1}{x^6+1}\,dx$$
Using (proof can be seen here)
$$\int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx=\frac{\pi}{b}\csc\left(\frac{a\pi}{b}\right)$$
then
$$I=\frac{1}{2}\int_{0}^{\infty}\frac{x^4+1}{x^6+1}\,dx=\frac{1}{2}\left[\frac{\pi}{6}\csc\left(\frac{5\pi}{6}\right)+\frac{\pi}{6}\csc\left(\frac{\pi}{6}\right)\right]=\frac{\pi}{3}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
&\color{#66f}{\large\int_{0}^{1}{x^{4} + 1 \over x^{6} + 1}\,\dd x}
=\half\pars{\int_{0}^{1}{x^{4} + 1 \over x^{6} + 1}\,\dd x
+\int_{0}^{1}{x^{4} + 1 \over x^{6} + 1}\,\dd x}
\\[5mm]&=\half\bracks{\int_{0}^{1}{x^{4} + 1 \over x^{6} + 1}\,\dd x
+\int_{\infty}^{1}{1/x^{4} + 1 \over 1/x^{6} + 1}\,\pars{-\,{\dd x \over x^{2}}}}
\\[5mm]&=\half\pars{\int_{0}^{1}{x^{4} + 1 \over x^{6} + 1}\,\dd x
+\int_{1}^{\infty}{x^{4} + 1 \over x^{6} + 1}\,\dd x}
=\half\int_{0}^{\infty}{x^{4} + 1 \over x^{6} + 1}\,\dd x
=\color{#66f}{\large%
{1 \over 4}\int_{-\infty}^{\infty}{x^{4} + 1 \over x^{6} + 1}\,\dd x}
\end{align}

Zeros of $\ds{\quad x^{6} + 1 = 0\quad}$ are given by
  $\ds{\quad x_{n} = \exp\pars{\bracks{2n + 1}\,{\pi \over 6}\,\ic}}$,
  $\ds{n = 0,1,2,\ldots,5}$, such that:

\begin{align}
&\color{#66f}{\large\int_{0}^{1}{x^{4} + 1 \over x^{6} + 1}\,\dd x}
={1 \over 4}\sum_{n\ =\ 0}^{2}2\pi\ic\lim_{x\ \to\ x_{n}}\bracks{%
\pars{x - x_{n}}\,{x^{4} + 1 \over x^{6} + 1}}
={\pi\ic \over 2}\sum_{n\ =\ 0}^{2}{x_{n}^{4} + 1 \over 6x_{n}^{5}}
\\[5mm]&={\pi\ic \over 12}
\sum_{n\ =\ 0}^{2}{x_{n}^{2} + x_{n}^{-2} \over x_{n}^{3}}
={\pi\ic \over 12}\sum_{n\ =\ 0}^{2}
{2\,\Re\pars{x_{n}^{2}} \over \pars{-1}^{n}\,\ic}
={\pi \over 6}\,\Re\sum_{n\ =\ 0}^{2}\pars{-1}^{n}x_{n}^{2}
\\[5mm]&={\pi \over 6}\,\Re\bracks{\exp\pars{{\pi \over 3}\,\ic}
-\exp\pars{\pi\ic} + \exp\pars{{5\pi \over 3}\,\ic}}
={\pi \over 6}\bracks{\half -\pars{-1} + \half}
=\color{#66f}{\large{\pi \over 3}}
\end{align}
