Question

\[\int\frac{x}{x^3 - 1} \text{ dx}\]

Answer 1

\[\text{We have}, \]
\[I = \int\frac{\text{ x dx}}{x^3 - 1}\]
\[ = \int\frac{\text{ x dx}}{\left( x - 1 \right) \left( x^2 + x + 1 \right)}\]
\[\text{ Let} \frac{x}{\left( x - 1 \right) \left( x^2 + x + 1 \right)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}\]
\[ \Rightarrow \frac{x}{\left( x - 1 \right) \left( x^2 + x + 1 \right)} = \frac{A \left( x^2 + x + 1 \right) + \left( Bx + C \right) \left( x - 1 \right)}{\left( x - 1 \right) \left( x^2 + x + 1 \right)}\]
\[ \Rightarrow x = A \left( x^2 + x + 1 \right) + B x^2 - Bx + Cx - C\]
\[ \Rightarrow x = \left( A + B \right) x^2 + \left( A - B + C \right) x + A - C\]
\[\text{Equating Coefficient of like terms}\]
\[A + B = 0 . . . . . \left( 1 \right)\]
\[A - B + C = 1 . . . . . \left( 2 \right)\]
\[A - C = 0 . . . . . \left( 3 \right)\]
\[\text{Solving} \left( 1 \right), \left( 2 \right) \text{ and }\left( 3 \right), \text{we get}\]
\[A = \frac{1}{3}\]
\[B = - \frac{1}{3}\]
\[C = \frac{1}{3}\]
\[ \therefore \frac{x}{\left( x - 1 \right) \left( x^2 + x + 1 \right)} = \frac{1}{3 \left( x - 1 \right)} + \frac{- \frac{1}{3}x + \frac{1}{3}}{x^2 + x + 1}\]
\[ = \frac{1}{3 \left( x - 1 \right)} + \frac{1}{3} \left[ \frac{- x + 1}{x^2 + x + 1} \right]\]
\[ = \frac{1}{3 \left( x - 1 \right)} - \frac{1}{3} \left[ \frac{x - 1}{x^2 + x + 1} \right]\]
\[ = \frac{1}{3 \left( x - 1 \right)} - \frac{1}{6} \left[ \frac{2x - 2}{x^2 + x + 1} \right]\]
\[ = \frac{1}{3 \left( x - 1 \right)} - \frac{1}{6} \left[ \frac{2x + 1}{x^2 + x + 1} \right] - \frac{1}{6} \times \frac{- 3}{x^2 + x + 1}\]
\[ = \frac{1}{3 \left( x - 1 \right)} - \frac{1}{6} \left[ \frac{2x + 1}{x^2 + x + 1} \right] + \frac{1}{2} \times \frac{1}{x^2 + x + 1}\]
\[ \therefore I = \frac{1}{3}\int\frac{dx}{x - 1} - \frac{1}{6}\int\frac{\left( 2x + 1 \right) dx}{x^2 + x + 1} + \frac{1}{2}\int\frac{dx}{x^2 + x + \frac{1}{4} - \frac{1}{4} + 1}\]
\[\text{ Putting x}^2 + x + 1 = t\]
\[ \Rightarrow \left( 2x + 1 \right) dx = dt\]
\[ \therefore I = \frac{1}{3} \text{ log }\left| x - 1 \right| - \frac{1}{6} \text{ log} \left| t \right| + \frac{1}{2}\int\frac{dx}{\left( x + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}\]
\[ = \frac{1}{3} \text{ log} \left| x - 1 \right| - \frac{1}{6} \text{ log} \left| x^2 + x + 1 \right| + \frac{1}{2}\left[ \frac{2}{\sqrt{3}} \text{ tan}^{- 1} \left( \frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) \right] + C\]
\[ = \frac{1}{3} \text{ log } \left| x - 1 \right| - \frac{1}{6} \text{ log} \left| x^2 + x + 1 \right| + \frac{1}{\sqrt{3}} \text{ tan}^{- 1} \left( \frac{2x + 1}{\sqrt{3}} \right) + C\]