# 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!

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

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 integrand actually doesn't change under your substitution, check your work. Also, the range ends out going from 1 to $\infty$ after the substitution. –  Zarrax Jan 21 '12 at 22:18
Your correction is also incorrect, as you are now subtracting $x^2$ twice from the numerator... change one of the $x^2$'s to a $-x^2$ and it should work. –  Zarrax Jan 21 '12 at 22:22
Edited, actually there was no need for the substitution... The previous answer was waaay more complicated than it needed :) –  N. S. Jan 21 '12 at 22:23
+1 Very clever! –  Américo Tavares Jan 21 '12 at 23:17
This answer, now, should be the accepted one (+1). –  John Bentin Jan 22 '12 at 15:10

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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