Question

How do you integrate $\int{\dfrac{1}{{{x}^{3}}-1}}$ using partial fractions?

Answer 1

Hint: Firstly use the partial fraction method to express the integrand in two separate integrands whose denominators are the individual factors of the given integrand. And then integrate them separately.
Partial fraction of the rational fraction form $\dfrac{p(x)+q}{(x\pm a)(x\pm b)}$ is given as $\dfrac{A}{(x\pm a)}+\dfrac{B}{(x\pm b)}$
Find the value of A and B by comparing it with original integrand.

Complete step by step solution:
In order to integrate $\int{\dfrac{1}{{{x}^{3}}-1}}$ using partial fractions, we will first separate or express the integrand as sum of two integrands as follows;
We can rewrite ${{x}^{3}}-1=\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)$using the identity.
$\dfrac{1}{{{x}^{3}}-1}=\dfrac{1}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}=\dfrac{A}{(x-1)}+\dfrac{Bx+C}{\left( {{x}^{2}}+x+1 \right)}$
Now, we will find the value of A,B and C by comparing both integrands as follows
\[\Rightarrow 1=A\left( {{x}^{2}}+x+1 \right)+\left( Bx+C \right)\left( x-1 \right)\]
Equating similar terms (constants and coefficients of the variable) we will get,
$\Rightarrow A+B=0\ \ then\ \Rightarrow A=-B$------ (1)
$\Rightarrow A-C=1\ \ then\ \Rightarrow C=A-1$------ (2)
$\Rightarrow A-B+C=0$------ (3)
Substituting equation (1) and equation (2) into equation (3),
$\Rightarrow A+A+A-1=0$
$\Rightarrow A=\dfrac{1}{3}$
$\Rightarrow B=-\dfrac{1}{3}$
$\Rightarrow C=\dfrac{1}{3}-1=\dfrac{1-3}{3}=-\dfrac{2}{3}$
Now we have the respective values of A, B and C, so putting them in the above considered integrands and then integrating them, we will get
$\dfrac{1}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}=\dfrac{1}{3(x-1)}+\dfrac{-\dfrac{1}{3}x-\dfrac{2}{3}}{\left( {{x}^{2}}+x+1 \right)}$
Simplifying the above expression, we get
$\dfrac{1}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}=\dfrac{1}{3(x-1)}-\dfrac{x+2}{3\left( {{x}^{2}}+x+1 \right)}$
Integrating the resultant expression, we obtain
$\Rightarrow \int{\dfrac{1}{3(x-1)}-\dfrac{x+2}{3\left( {{x}^{2}}+x+1 \right)}}dx$
Distributing the integration above subtraction, we will get
$\Rightarrow \int{\dfrac{1}{3(x-1)}-\dfrac{x+2}{3\left( {{x}^{2}}+x+1 \right)}}dx=\int{\dfrac{1}{3(x-1)}dx-\int{\dfrac{x+2}{3\left( {{x}^{2}}+x+1 \right)}dx}}$
Now, we know that the integration of $\dfrac{1}{(x\pm a)}$ equals $\ln \left| x\pm a \right|+c$, so integrating the above functions using this formula, we will get
$\Rightarrow \int{\dfrac{1}{3(x-1)}dx-\int{\dfrac{x+2}{3\left( {{x}^{2}}+x+1 \right)}dx}}=\dfrac{1}{3}\ln \left| x-1 \right|+{{C}_{1}}-\dfrac{1}{3}\int{\dfrac{x+2}{\left( {{x}^{2}}+x+1 \right)}dx}$
We can rewrite this expression $\dfrac{x+2}{\left( {{x}^{2}}+x+1 \right)}$ as \[\dfrac{1}{2}\left( \dfrac{2x+1}{{{x}^{2}}+x+1}+\dfrac{3}{{{x}^{2}}+x+1} \right)\]
Thus, substituting this value, we obtain
$\Rightarrow \int{\dfrac{1}{3(x-1)}dx-\int{\dfrac{x+2}{3\left( {{x}^{2}}+x+1 \right)}dx}}=\dfrac{1}{3}\ln \left| x-1 \right|+{{C}_{1}}-\dfrac{1}{3}\int{\dfrac{1}{2}\left( \dfrac{2x+1}{{{x}^{2}}+x+1}+\dfrac{3}{{{x}^{2}}+x+1} \right)dx}$
Simplifying the above expression, we get
$\Rightarrow \int{\dfrac{1}{3(x-1)}dx-\int{\dfrac{x+2}{3\left( {{x}^{2}}+x+1 \right)}dx}}=\dfrac{1}{3}\ln \left| x-1 \right|+{{C}_{1}}-\dfrac{1}{6}\int{\left( \dfrac{2x+1}{{{x}^{2}}+x+1}+\dfrac{3}{{{x}^{2}}+x+1} \right)dx}$
Now solving,
$\Rightarrow -\dfrac{1}{6}\int{\left( \dfrac{2x+1}{{{x}^{2}}+x+1}+\dfrac{3}{{{x}^{2}}+x+1} \right)dx}=-\dfrac{1}{6}\ln \left| {{x}^{2}}+x+1 \right|+{{C}_{2}}-\dfrac{1}{6}\int{\dfrac{3}{{{x}^{2}}+x+1}}dx$
Simplifying the above by splitting middle term of the quadratic equation.
As we know that,
$\Rightarrow -\dfrac{1}{6}\int{\dfrac{3}{{{x}^{2}}+x+1}}dx=-\dfrac{3}{6}\int{\dfrac{1}{{{\left( x+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}}dx}=-\dfrac{1}{2}\int{\dfrac{1}{{{\left( x+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}}dx}$
Let substitute\[s=x+\dfrac{1}{2}\ then\ ds=dx\],
$\Rightarrow -\dfrac{1}{2}\int{\dfrac{1}{{{\left( x+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}}dx}=-\dfrac{1}{2}\int{\dfrac{1}{{{s}^{2}}+\dfrac{3}{4}}dx}==-\dfrac{1}{2}\int{\dfrac{\dfrac{4}{3}}{\dfrac{4}{3}\left( {{s}^{2}}+\dfrac{3}{4} \right)}dx=-\dfrac{2}{3}\int{\dfrac{1}{\dfrac{4}{3}{{s}^{2}}+1}}}dx$
Let \[p=\dfrac{2}{\sqrt{3}}s\ then\ dp=\dfrac{2}{\sqrt{3}}ds\]
Therefore,
$\Rightarrow -\dfrac{2}{3}\int{\dfrac{1}{\dfrac{4}{3}{{s}^{2}}+1}}ds=-\dfrac{1}{\sqrt{3}}\int{\dfrac{1}{{{p}^{2}}+1}dp}$
The given integral is an arctangent integral.
Thus,
$=-\dfrac{1}{\sqrt{3}}\int{\dfrac{1}{{{p}^{2}}+1}dp}=-\dfrac{1}{\sqrt{3}}\arctan p+{{C}_{3}}$
Undo the substitution, we get
$\Rightarrow -\dfrac{1}{\sqrt{3}}\arctan p+{{C}_{3}}=-\dfrac{\sqrt{3}}{3}\arctan \left( \dfrac{\left( 2x+1 \right)\sqrt{3}}{3} \right)+{{C}_{3}}$
Now,
Combining all three integrals, we get
$\Rightarrow \int{\dfrac{1}{{{x}^{3}}-1}}=\dfrac{1}{3}\ln \left| x-1 \right|-\dfrac{1}{6}\ln \left| {{x}^{2}}+x+1 \right|-\dfrac{\sqrt{3}}{3}\arctan \left( \dfrac{\left( 2x+1 \right)\sqrt{3}}{3} \right)+C$
Taking out \[\dfrac{1}{6}\] as a common factor, we get
$\Rightarrow \int{\dfrac{1}{{{x}^{3}}-1}}=\dfrac{1}{6}\left( 2\ln \left| x-1 \right|-\ln \left| {{x}^{2}}+x+1 \right|-2\sqrt{3}\arctan \left( \dfrac{\left( 2x+1 \right)\sqrt{3}}{3} \right) \right)+C$
Hence, it is the required integration.

Note: When doing indefinite integration, always write $+c$ part after the integration. This $+c$ part indicates the constant part remains after integration and can be understood when you explore it graphically. Infinite integration constant gets cancelled out, so we only write it in indefinite integration.