Question

Find integral solution of the following expression$\dfrac{1}{{{x^3} - 1}}$using the partial fraction method.

Answer 1

In this question, we need to find an integral solution of the expression $\dfrac{1}{{{x^3} - 1}}$ with respect to x, using the partial fraction method. In algebra, partial fraction method or partial decomposition method is a method under which a rational fraction is expressed as the sum of a polynomial or one or more fractions with simpler denominator.

Complete step by step solution:
We have,
$\Rightarrow \dfrac{1}{{{x^3} - 1}}$
Considering denominator of the above fraction, that is,
$\Rightarrow {x^3} - 1$
Here,we know that,
${1^3} = 1$
Therefore, we can rewrite the given fraction as,
$\Rightarrow \dfrac{1}{{{x^3} - {1^3}}}$
We also know that, ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$
Now, given is a polynomial of power three, it can be reduced to factors of two polynomial using the above identity as shown below,
$\Rightarrow \dfrac{1}{{{x^3} - {1^3}}} = \dfrac{1}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} - - - - - \left( 1 \right)$
Now, according to the partial fraction method we need to change the right hand side equation into a sum of terms that can be integrated easily.
Thus.
$\Rightarrow \dfrac{1}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{A}{{\left( {x - 1} \right)}} + \dfrac{{Bx + C}}{{{x^2} + x + 1}}$
Taking LCM,
$\Rightarrow \dfrac{{A\left( {{x^2} + x + 1} \right) + (Bx + C)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}$
Here, we solve the following equation to find the values of A, B and C,
$\Rightarrow A\left( {{x^2} + x + 1} \right) + (Bx + C)\left( {x - 1} \right) = 1$
Let, $x = 1$
Therefore,
$\Rightarrow A\left( {1 + 1 + 1} \right) + \left( {B + C} \right)\left( 0 \right) = 1$
$\Rightarrow A = \dfrac{1}{3}$
Now, let
$x = 0$
Therefore we get,
$\Rightarrow A\left( {0 + 0 + 1} \right) + \left( {0 + C} \right)\left( { - 1} \right) = 1$
$\Rightarrow C = \dfrac{{ - 2}}{3}$
Finally let,
$x = - 1$
Therefore,
$\Rightarrow A\left( {1 - 1 + 1} \right) + \left( { - B + C} \right)\left( { - 2} \right) = 1$
$\Rightarrow B = \dfrac{{ - 1}}{2}$
Now, putting the values of A,B and C in (1)
Thus,
$\Rightarrow \dfrac{1}{{{x^3} - 1}} = \dfrac{1}{3} \times \dfrac{1}{{\left( {x - 1} \right)}} + \dfrac{{\dfrac{{ - 1}}{3}x - \dfrac{2}{3}}}{{\left( {{x^2} + x + 1} \right)}}$
Taking integration, we get
$\Rightarrow \dfrac{1}{3}\int {\dfrac{1}{{\left( {x - 1} \right)}}dx - } \dfrac{1}{3}\int {\dfrac{{\left( {x + 2} \right)}}{{\left( {{x^2} + x + 1} \right)}}dx}$
Here, we know
$\Rightarrow \dfrac{1}{3}\int {\dfrac{1}{{\left( {x - 1} \right)}}dx = \dfrac{1}{3}\log \left| {x - 1} \right|}$
Thus,
$\Rightarrow \dfrac{1}{3}\int {\dfrac{1}{{\left( {x - 1} \right)}}dx - } \dfrac{1}{3}\int {\dfrac{{\left( {x + 2} \right)}}{{\left( {{x^2} + x + 1} \right)}}dx} = \dfrac{1}{3}\log \left| {x - 1} \right| - \dfrac{1}{3}\int {\dfrac{{\left( {x + 2} \right)}}{{\left( {{x^2} + x + 1} \right)}}dx}$
Now, when we multiply and divide 2 to the second part of the equation we get,
$\Rightarrow \dfrac{1}{3}\int {\dfrac{{\left( {x + 2} \right)}}{{\left( {{x^2} + x + 1} \right)}}dx} = \dfrac{1}{6}\int {\dfrac{{2x + 4}}{{{x^2} + x + 1}}dx}$
Thus, we get the whole equation as,
$\Rightarrow \dfrac{1}{3}\log \left| {x - 1} \right| - \dfrac{1}{6}\int {\dfrac{{2x + 4}}{{{x^2} + x + 1}}dx}$
Which can b rewritten as,
$\Rightarrow \dfrac{1}{3}\log \left| {x - 1} \right| - \dfrac{1}{6}\int {\dfrac{{2x + 1 + 3}}{{{x^2} + x + 1}}} dx$
Therefore, separating the above equation, we get,
$\Rightarrow \dfrac{1}{3}\log \left| {x - 1} \right| - \dfrac{1}{6}\int {\dfrac{{2x + 1}}{{{x^2} + x + 1}}} dx + \dfrac{3}{6}\int {\dfrac{1}{{{x^2} + x + 1}}} dx$
Solving integral for the second part of the above equation,
$\Rightarrow \dfrac{1}{6}\int {\dfrac{{2x + 1}}{{{x^2} + x + 1}}} dx = \dfrac{1}{6}\log \left| {{x^2} + x + 1} \right|$
Thus we get the whole equation as,
$\Rightarrow \dfrac{1}{3}\log \left| {x - 1} \right| - \dfrac{1}{6}\log \left| {{x^2} + x + 1} \right| + \dfrac{1}{2}\int {\dfrac{1}{{{x^2} + x + 1}}} dx$
Now, for the third part of the above equation,
We know that, $1 = \dfrac{1}{4} + \dfrac{3}{4}$
Therefore,
We can rewrite the third part of the equation, which is
$\dfrac{1}{2}\int {\dfrac{1}{{{x^2} + x + 1}}} dx$ as
$\dfrac{1}{2}\int {\dfrac{1}{{\left( {{x^2} + x + \dfrac{1}{4} + \dfrac{3}{4}} \right)}}} dx$
Here, we know that, ${x^2} + {\left( {\dfrac{1}{2}} \right)^2} = {x^2} + x + \dfrac{1}{4}$
And ${\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} = \dfrac{3}{4}$
Therefore,
$\Rightarrow \dfrac{1}{2}\int {\dfrac{1}{{\left( {{x^2} + x + \dfrac{1}{4} + \dfrac{3}{4}} \right)}}} dx = \dfrac{1}{2}\int {\dfrac{1}{{{{\left( {x + \dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}}}} dx$
Thus, according to the integration identity,
We know that $\dfrac{1}{2}\int {\dfrac{1}{{{{\left( {x + \dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}}}} dx = \dfrac{1}{2} \times \dfrac{1}{{\dfrac{{\sqrt 3 }}{2}}}{\tan ^{ - 1}}\left( {\dfrac{{x + \dfrac{1}{2}}}{{\dfrac{{\sqrt 3 }}{2}}}} \right) + C$
Thus, our solution becomes

$\Rightarrow \dfrac{1}{3}\log \left| {x - 1} \right| - \dfrac{1}{6}\log \left| {{x^2} + x + 1} \right| - \dfrac{1}{2} \times \dfrac{1}{{\dfrac{{\sqrt 3 }}{2}}}{\tan ^{ - 1}}\left( {\dfrac{{2x + 1}}{{\sqrt 3 }}} \right) + C$
Which is our required integral.

Note:
It is very important to note here that the partial fraction method can only be applied to fractions whose degree of denominator is strictly more than that of the numerator. Also,to solve the questions through partial integration methods, it is very important to remember the standard integrals.