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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!

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  • $\begingroup$ $\mathrm{d}x$ went missing! $\endgroup$ – user21436 Jan 21 '12 at 17:58
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    $\begingroup$ W.A. says that you should use partial fractions... $\endgroup$ – Peđa Terzić Jan 21 '12 at 18:00
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    $\begingroup$ W.A. says it is $\pi/3$. $\endgroup$ – Jon Jan 21 '12 at 18:16
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    $\begingroup$ I'm trying to find a way for computing it easily without using W.A... $\endgroup$ – Jozef Jan 21 '12 at 18:22
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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...

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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 $\arctan \dfrac{4}{1-1}= \dfrac{\pi }{2}.$

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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 $$ I = \frac{\pi}{3} $$

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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.

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one way is partial fractions on $$ \frac{x^4+1}{x^6+1}=\frac{(x-e^{\pi i/4})(x-e^{3\pi i/4})(x-e^{5\pi i/4})(x-e^{7\pi i/4})}{(x-e^{\pi i/6})(x-e^{3\pi i/6})(x-e^{5\pi i/6})(x-e^{7\pi i/6})(x-e^{9\pi i/6})(x-e^{11\pi i/6})} $$ $$ =\frac{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}{(x^2+1)(x^2+\sqrt{3}x+1)(x^2-\sqrt{3}x+1)} $$

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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 $$I=\frac{1}{2}\int_{0}^{\infty}\frac{x^4+1}{x^6+1}\,dx=\frac{1}{2}\left[\frac{\pi}{6}\csc\left(\frac{5\pi}{6}\right)+\frac{\pi}{6}\csc\left(\frac{\pi}{6}\right)\right]=\frac{\pi}{3}$$

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}

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