Question

\[\int\frac{2x}{x^3 - 1} dx\]

Answer 1

We have,
\[I = \int\frac{2x dx}{x^3 - 1}\]
\[ = \int\frac{2x dx}{\left( x - 1 \right) \left( x^2 + x + 1 \right)}\]
\[\text{Let }\frac{2x}{\left( x - 1 \right) \left( x^2 + x + 1 \right)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}\]
\[ \Rightarrow \frac{2x}{\left( x - 1 \right) \left( x^2 + x + 1 \right)} = \frac{A \left( x^2 + x + 1 \right) + \left( Bx + C \right) \left( x - 1 \right)}{\left( x - 1 \right) \left( x^2 + x + 1 \right)}\]
\[ \Rightarrow 2x = A \left( x^2 + x + 1 \right) + B x^2 - Bx + Cx - C\]
\[ \Rightarrow 2x = \left( A + B \right) x^2 + \left( A - B + C \right) x + A - C\]
\[\text{Equating coefficients of like terms}\]
\[A + B = 0 . . . . . \left( 1 \right)\]
\[A - B + C = 2 . . . . . \left( 2 \right)\]
\[A - C = 0 . . . . . \left( 3 \right)\]
\[\text{Solving (1), (2) and (3), we get}\]
\[A = \frac{2}{3}\]
\[B = - \frac{2}{3}\]
\[C = \frac{2}{3}\]
\[ \therefore I = \frac{2}{3}\int\frac{dx}{x - 1} + \frac{2}{3}\int\left( \frac{- x + 1}{x^2 + x + 1} \right)dx\]
\[\text{Let }- x + 1 = a\frac{d}{dx} \left( x^2 + x + 1 \right) + b\]
\[ \Rightarrow - x + 1 = a \left( 2x + 1 \right) + b\]
\[ \Rightarrow - x + 1 = \left( 2a \right) x + a + b\]
\[\text{Equating coefficients of like terms}\]
\[ 2a = - 1\]
\[ \Rightarrow a = - \frac{1}{2}\]
And
\[a + b = 1\]
\[ \Rightarrow - \frac{1}{2} + b = 1\]
\[ \Rightarrow b = \frac{3}{2}\]
\[ \therefore I = \frac{2}{3}\int\frac{dx}{x - 1} + \frac{2}{3}\int\left[ \frac{- \frac{1}{2} \left( 2x + 1 \right) + \frac{3}{2}}{\left( x^2 + x + 1 \right)} \right]dx\]
\[ = \frac{2}{3}\int\frac{dx}{x - 1} - \frac{1}{3}\int\left( \frac{2x + 1}{x^2 + x + 1} \right) dx + \int\frac{dx}{x^2 + x + 1}\]
\[ = \frac{2}{3}\int\frac{dx}{x - 1} - \frac{1}{3}\int\left( \frac{2x + 1}{x^2 + x + 1} \right)dx + \int\frac{dx}{x^2 + x + \frac{1}{4} - \frac{1}{4} + 1}\]
\[ = \frac{2}{3}\int\frac{dx}{x - 1} - \frac{1}{3}\int\frac{\left( 2x + 1 \right) dx}{x^2 + x + 1} + \int\frac{dx}{\left( x + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}\]
\[\text{Let }x^2 + x + 1 = t\]
\[ \Rightarrow \left( 2x + 1 \right) dx = dt\]
Then,
\[I = \frac{2}{3}\int\frac{dx}{x - 1} - \frac{1}{3}\int\frac{dt}{t} + \int\frac{dx}{\left( x + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}\]
\[ = \frac{2}{3} \log \left| x - 1 \right| - \frac{1}{3} \log \left| t \right| + \frac{2}{\sqrt{3}} \tan^{- 1} \left( \frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) + C'\]
\[ = \frac{2}{3} \log \left| x - 1 \right| - \frac{1}{3} \log \left| x^2 + x + 1 \right| + \frac{2}{\sqrt{3}} \tan^{- 1} \left( \frac{2x + 1}{\sqrt{3}} \right) + C'\]