# Thread:

1. ## a baffling integral

∫ 1 / (x^6 – 1) dx

Thanks in advance.

2. Originally Posted by niz

∫ 1 / (x^6 – 1) dx

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

You should be able to apply the method of partial fractions for the case when the denominator contains irreducible quadratic factors.

3. Here's my method , it is for those who HATE the method of partial fraction ( but i am not the one )

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

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

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

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

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

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

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

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

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

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

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

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

the integral becomes

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

now consider

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

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

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

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

and

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

$(2) - (1)$ ,

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

therefore ,

$\int \frac{dx}{x^6- 1} = \frac{1}{6}\ln{ \left( \frac{ x^3 - 1}{x^3 + 1} \right ) } - \frac{1}{2}[ \frac{1}{\sqrt{3}} \tan^{-1}(\frac{x^2-1}{\sqrt{3} x } ) - \frac{1}{2} \ln{ \left( \frac{ x^2 - x +1}{x^2 + x + 1} \right ) } ] + C$

4. ## 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.
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