# a baffling integral

• Dec 20th 2009, 08:39 PM
niz
a baffling integral

∫ 1 / (x^6 – 1) dx

• Dec 20th 2009, 10:29 PM
Quote:

Originally Posted by niz

∫ 1 / (x^6 – 1) dx

Are you familiar with the method of partial fractions? Notice that the denominator is the difference of two squares $\displaystyle (x^3-1)(x^3+1)$. Now use the formulae for the cubes. You can write that out as $\displaystyle (x-1)(x^2+x+1)(x+1)(x^2-x+1)$

You should be able to apply the method of partial fractions for the case when the denominator contains irreducible quadratic factors.
• Dec 21st 2009, 02:38 AM
simplependulum
Here's my method , it is for those who HATE the method of partial fraction ( but i am not the one )

$\displaystyle \frac{1}{x^6- 1} = \frac{ x^2 - ( x^2 -1 )}{ x^6 - 1}$

$\displaystyle = \frac{x^2}{x^6-1} - \frac{ x^2 - 1}{ ( x^2 - 1 )( x^4 + x^2 + 1 )}$

$\displaystyle = \frac{x^2}{x^6-1} - \frac{1}{ x^4 + x^2 + 1}$

the integral $\displaystyle \int \frac{dx}{x^6 - 1} = \int \frac{x^2 dx}{x^6-1} - \int \frac{dx}{ x^4 + x^2 + 1}$

the first one we just need to substitute $\displaystyle t = x^3$ and finally obtain

$\displaystyle \frac{1}{6}\ln{ \left( \frac{ x^3 - 1}{x^3 + 1} \right ) }$

you may think that we may need to apply partial fraction to solve the second part but if we consider

$\displaystyle \int \frac{ x^2 + 1}{ x^4 + x^2 + 1}~dx$

Divide the numerator and the denominator by $\displaystyle x^2$

$\displaystyle = \int \frac{ 1 + 1/x^2}{ x^2 + 1 + 1/x^2 } ~dx$

$\displaystyle = \int \frac{ 1 + 1/x^2}{ \left ( x - \frac{1}{x} \right )^2 + 3 }$

then substitute $\displaystyle x- \frac{1}{x} = t , (1 + 1/x^2 )dx = dt$

the integral becomes

$\displaystyle \int \frac{dt}{ t^2 + 3 } = \frac{1}{\sqrt{3}} \tan^{-1}(\frac{x^2-1}{\sqrt{3} x } )$

now consider

$\displaystyle \int \frac{x^2 - 1 }{ x^4 + x^2 + 1}~dx$

do the same thing above but sub. $\displaystyle x + 1/x = t$ this time , we can get

$\displaystyle = \frac{1}{2} \ln{ \left( \frac{ x^2 - x +1}{x^2 + x + 1} \right ) }$

we have
$\displaystyle \int \frac{ x^2 - 1}{ x^4 + x^2 + 1}~dx = \frac{1}{2} \ln{ \left( \frac{ x^2 - x +1}{x^2 + x + 1} \right ) }$ $\displaystyle (1)$

and

$\displaystyle \int \frac{ x^2 + 1}{ x^4 + x^2 + 1}~dx = \frac{1}{\sqrt{3}} \tan^{-1}(\frac{x^2-1}{\sqrt{3} x } )$ $\displaystyle (2)$

$\displaystyle (2) - (1)$ ,

$\displaystyle \int \frac{dx}{ x^4 + x^2 + 1} = \frac{1}{2}[ \frac{1}{\sqrt{3}} \tan^{-1}(\frac{x^2-1}{\sqrt{3} x } ) - \frac{1}{2} \ln{ \left( \frac{ x^2 - x +1}{x^2 + x + 1} \right ) } ]$

therefore ,

$\displaystyle \int \frac{dx}{x^6- 1} = \frac{1}{6}\ln{ \left( \frac{ x^3 - 1}{x^3 + 1} \right ) } - \frac{1}{2}[ \frac{1}{\sqrt{3}} \tan^{-1}(\frac{x^2-1}{\sqrt{3} x } ) - \frac{1}{2} \ln{ \left( \frac{ x^2 - x +1}{x^2 + x + 1} \right ) } ] + C$
• Dec 22nd 2009, 07:31 AM
niz
Thanks
Thanks adkinsjr and simplependulum!! Actually I was hoping for some kind of trigonometric substitution to manipulate the integrand. Like for (x^2 -1) substituting x= secy etc.. I guess there are none such solutions right? Anyway, thanks guys.. Thanks simplependulum for the smart manipulation of the integrand..that was clever.