Integrating a Rational Function Whose Denominator Contains a Distinct Irreducible Quadratic Factor

Find \(\int \dfrac{3x}{x^{3}-1}\,dx\).

Solution  The denominator is \(x^{3}-1=(x-1)(x^{2}+x+1)\). Since \( x-1\) is a nonrepeated linear factor, by Case 1 the decomposition of \(\dfrac{ 3x}{x^{3}-1}=\dfrac{3x}{(x-1) (x^{2}+x+1) }\) has the term \(\dfrac{A}{x-1}\).

The discriminant of the quadratic equation \(x^{2}+x+1=0\) is negative, so \( x^{2}+x+1\) is an irreducible quadratic factor of \(q\), and the decomposition of \(\dfrac{p}{q}\) also contains the term \(\dfrac{Bx+C}{x^{2}+x+1}\). Then \[ \dfrac{3x}{x^{3}-1}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^{2}+x+1} \]

Clearing the denominators, we have \[ \begin{equation*} 3x=A(x^{2}+x+1)+(Bx+C)(x-1)\ \end{equation*} \]

This is an identity in \(x\). When \(x=1\), we have \(3=3A,\) so \(A=1.\) With \(A=1,\) the identity becomes \[ \begin{eqnarray*} 3x &=& (x^{2}+x+1) + (Bx+C) (x-1) \\[3pt] -x^{2}+2x-1 &=& (Bx+C) (x-1) \\[3pt] -(x-1) ^{2} &=& (Bx+C) (x-1) \\[3pt] -(x-1) &=& Bx+C \\[3pt] B &=& -1,\quad C=1 \end{eqnarray*} \]

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So, \[ \begin{eqnarray*} \dfrac{3x}{x^{3}-1} &=&\dfrac{1}{x-1}+\dfrac{-x+1}{x^{2}+x+1} \nonumber\\[5pt] \int \dfrac{3x}{x^{3}-1}\,dx &=&\int \left( \dfrac{1}{x-1}+\dfrac{-x+1}{ x^{2}+x+1}\right) dx=\int \dfrac{1}{x-1}\,dx+\int \dfrac{-x+1}{x^{2}+x+1}\,dx \nonumber\\[5pt] &=& \ln \vert x-1\vert -\int \dfrac{x-1}{x^{2}+x+1}\, dx\tag{2} \end{eqnarray*} \]

To find the integral on the right, we complete the square in the denominator and use substitution. \[ \begin{eqnarray*} \int \dfrac{x-1}{x^{2}+x+1}dx &=& \int \dfrac{x-1}{\left( x+\dfrac{1}{2}\right)^{2}+\dfrac{3}{4}}dx \underset{\underset{\color{#0066A7}{\hbox{\(u=x+\dfrac{1}{2}\)}}}{\color{#0066A7}{\left\uparrow{\vphantom{\vrule width0pc height11pt\\ depth0pt}}\right.}}} {=} \int \dfrac{u-\dfrac{3}{2}}{u^{2}+\dfrac{3}{4}}du=\int \dfrac{u}{u^{2}+\dfrac{3}{4\\ }}du-\dfrac{3}{2}\int \dfrac{du}{u^2+\dfrac{3}{4}} \\ &=& \dfrac{1}{2}\ln \left( u^{2}+\dfrac{3}{4}\right) -\dfrac{3}{2}\left[ \dfrac{2}{\sqrt{3}}\tan ^{-1}\left( \dfrac{2}{\sqrt{3}}u\right) \right] \quad {\color{#0066A7}{\hbox{\(\int \dfrac{du}{u^{2}+a^{2}}\, =\dfrac{1}{a}\tan^{-1}\left( \dfrac{u}{a}\right)\)}}} \\ &=& \dfrac{1}{2}\ln \left( u^{2}+\dfrac{3}{4}\right) -\sqrt{3}\tan ^{-1}\left( \dfrac{2}{\sqrt{3}}u\right) \\ &=& \dfrac{1}{2}\ln (x^{2}+x+1) -\sqrt{3}\tan ^{-1}\left( \dfrac{2x+1}{\sqrt{3}}\right) \end{eqnarray*} \]

Then from (2), \[ \int \dfrac{3x}{x^{3}-1}dx=\ln \vert x-1\vert -\dfrac{1}{2}\ln \,(x^{2}+x+1)+\sqrt{3}\tan ^{-1}\left( \dfrac{2x+1}{\sqrt{3}}\right) +C_{1} \]