# Use partial fractions to find \int \frac{1}{x^6 - 1}\ dx

## Question:

Use partial fractions to find

{eq}\int \frac{1}{x^6 - 1}\ dx {/eq}

## Partial fraction:

While using Partial Fraction in integration, an operation that consists of expressing the fraction as a sum of one or several fractions with a simpler denominator. According to this method, it is always possible to write the integrand as a sum of simpler rational functions by a method of partial fraction decomposition.

## Answer and Explanation:

we have,

{eq}\int {\frac{1}{{{x^6} - 1}}dx} {/eq}

Using Partial Fraction Decomposition we get the following equation,

{eq}\frac{1}{{\left( {x + 1} \right)\left( {x - 1} \right)\left( {{x^4} + {x^2} + 1} \right)}} = \frac{{{a_0}}}{{x + 1}} + \frac{{{a_1}}}{{x - 1}} + \frac{{{a_5}{x^3} + {a_4}{x^2} + {a_3}x + {a_2}}}{{{x^4} + {x^2} + 1}} {/eq}

{eq}1 = {a_0}\left( {x - 1} \right)\left( {{x^4} + {x^2} + 1} \right) + {a_1}\left( {x + 1} \right)\left( {{x^4} + {x^2} + 1} \right) + \left( {{a_5}{x^3} + {a_4}{x^2} + {a_3}x + {a_2}} \right)\left( {x + 1} \right)\left( {x - 1} \right) {/eq}

For the denominator values -1 and 1, we have {eq}a_0=-\frac{1}{6} {/eq} and {eq}a_1 = \frac{1}{6} {/eq} respectively.

Again by expanding and extracting the variables within the fraction we have,

{eq}1=a_5x^5+a_4x^4+\frac{1}{3}x^4+\frac{1}{3}x^2+\frac{1}{3}+a_3x^3-a_5x^3+a_2x^2-a_4x^2-a_3x-a_2 {/eq}

Now equating the coefficients of similar terms on both side,

{eq}\left[ {\begin{array}{*{20}{c}} {\frac{1}{3} - {a_2} = 1} \\ { - {a_3} = 0} \\ {\frac{1}{3} + {a_2} - {a_4} = 0} \\ {{a_3} - {a_5} = 0} \\ {{a_4} + \frac{1}{3} = 0} \\ {{a_5} = 0} \end{array}} \right] {/eq}

we get

{eq}a_4=-\frac{1}{3},\:a_3=0,\:a_2=-\frac{2}{3},\:a_5=0 {/eq}

Now putting all these solutions to the partial fraction parameters we get,

{eq}\frac{{ - {x^2} - 2}}{{3\left( {{x^4} + {x^2} + 1} \right)}} + \frac{1}{{6\left( {x - 1} \right)}} - \frac{1}{{6\left( {x + 1} \right)}} {/eq}

Now we have to integrate the above equation to get the solution to the given problem.

{eq}\int {\left( {\frac{{ - {x^2} - 2}}{{3\left( {{x^4} + {x^2} + 1} \right)}} + \frac{1}{{6\left( {x - 1} \right)}} - \frac{1}{{6\left( {x + 1} \right)}}} \right)} dx {/eq}

Solving the three terms separately

(i) {eq}\int {\frac{1}{{6(x - 1)}}dx = \frac{1}{6}\log (x - 1)} {/eq}

(ii) {eq}- \int {\frac{1}{{6(x + 1)}}dx = - \frac{1}{6}\log (x + 1)} {/eq}

(iii) {eq}\int {\left( {\frac{{ - {x^2} - 2}}{{3\left( {{x^4} + {x^2} + 1} \right)}}} \right)dx = - \frac{1}{3}\int {\left( {\frac{{{x^2} + 2}}{{{x^4} + {x^2} + 1}}} \right) dx= - \frac{1}{3}\int {\left( {\frac{{\left( {{x^2} + 1} \right) - \left( {{x^2} - 1} \right)}}{{{x^4} + {x^2} + 1}}} \right)dx} } } {/eq}

{eq}- \frac{1}{3}\int {\left( {\frac{{\left( {{x^2} + 1} \right) - \left( {{x^2} - 1} \right)}}{{{x^4} + {x^2} + 1}}} \right)dx} = - \frac{1}{3}\int {\frac{{{x^2} + 1}}{{{x^4} + {x^2} + 1}}}dx - \frac{1}{3}\int {\frac{{1 - {x^2}}}{{{x^4} + {x^2} + 1}}}dx {/eq}

{eq}- \frac{1}{3}\int {\frac{{{x^2} + 1}}{{{x^4} + {x^2} + 1}}} dx = - \frac{1}{3}\int {\frac{{1 + \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}} + 1}}dx = - \frac{1}{3}\int {\frac{{1 + \frac{1}{{{x^2}}}}}{{{{\left( {x - \frac{1}{x}} \right)}^2} + 3}}} } dx {/eq}

by putting {eq}x-\frac{1}{x} {/eq} as t and differentiating it with respect to x then we get the numerator which cancels out each other resulting in

{eq}\begin{gathered} - \frac{1}{3}\int {\frac{{1 + \frac{1}{{{x^2}}}}}{{{{\left( {x - \frac{1}{x}} \right)}^2} + 3}}} dx = - \frac{1}{3}\int {\frac{1}{{{{\left( {\sqrt 3 } \right)}^2} + {t^2}}}dt = } - \frac{1}{3}\left( {\frac{1}{{\sqrt 3 }}{{\tan }^{ - 1}}\frac{t}{{\sqrt 3 }}} \right) \hfill \\ - \frac{1}{3}\left( {\frac{1}{{\sqrt 3 }}{{\tan }^{ - 1}}\frac{{{x^2} - 1}}{{\sqrt 3 x}}} \right) \hfill \\ \end{gathered} {/eq}

Also for the second part

{eq}\begin{gathered} - \frac{1}{3}\int {\frac{{1 - \frac{1}{{{x^2}}}}}{{{{\left( {x + \frac{1}{x}} \right)}^2} - 1}}dx = - \frac{1}{3}\int {\frac{1}{{{t^2} - 1}}dt} } \hfill \\ - \frac{1}{3}\left( {\frac{1}{2}\log \left| {\frac{{t - 1}}{{t + 1}}} \right|} \right) = - \frac{1}{6}\left( {\log \left| {\frac{{{x^2} - x + 1}}{{{x^2} + x + 1}}} \right|} \right) \hfill \\ \end{gathered} {/eq}

So, the solution for the given problem is {eq}\frac{1}{6}\log \left| {\frac{{x - 1}}{{x + 1}}} \right| - \frac{1}{{3\sqrt 3 }}{\tan ^{ - 1}}\frac{{{x^2} - 1}}{{\sqrt 3 x}} - \frac{1}{6}\log \left| {\frac{{{x^2} - x + 1}}{{{x^2} + x + 1}}} \right| + c. {/eq}