I am baffled in solving the following integration problem. Please help!!
∫ 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 $\displaystyle (x^3-1)(x^3+1)$. 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.
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 $
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.