Question

\[\int\frac{x - 1}{3 x^2 - 4x + 3} dx\]

Answer 1

\[\int\left( \frac{x - 1}{3 x^2 - 4x + 3} \right)dx\]
\[x - 1 = A\frac{d}{dx}\left( 3 x^2 - 4x + 3 \right) + B\]
\[x - 1 = A \left( 6x - 4 \right) + B\]
\[x - 1 = \left( 6 A \right) x + B - 4 A\]

Comparing the Coefficients of like powers of x

\[\text{6 } A = 1\]
\[A = \frac{1}{6}\]
\[B - \text{ 4 A }= - 1\]
\[B - 4 \times \frac{1}{6} = - 1\]
\[B = - 1 + \frac{2}{3}\]
\[B = \frac{1}{3}\]

\[Now, \int\frac{\left( x - 1 \right) dx}{3 x^2 - 4x + 3}\]
\[ = \int\left[ \frac{\frac{1}{6}\left( 6x - 4 \right) + \frac{1}{3}}{3 x^2 - 4x + 3} \right]dx\]
\[ = \frac{1}{6}\int\frac{\left( 6x - 4 \right) dx}{3 x^2 - 4x + 3} + \frac{1}{3}\int\frac{dx}{3 x^2 - 4x + 3}\]
\[ = \frac{1}{6}\int\frac{\left( 6x - 4 \right) dx}{3 x^2 - 4x + 3} + \frac{1}{9}\int\frac{dx}{x^2 - \frac{4}{3}x + 1}\]
\[ = \frac{1}{6}\int\frac{\left( 6x - 4 \right) dx}{3 x^2 - 4x + 3} + \frac{1}{9}\int\frac{dx}{x^2 - \frac{4}{3}x + \left( \frac{2}{3} \right)^2 \left( \frac{2}{3} \right)^2 + 1}\]
\[ = \frac{1}{6}\int\frac{\left( 6x - 4 \right) dx}{3 x^2 - 4x + 3} + \frac{1}{9}\int\frac{dx}{\left( x - \frac{2}{3} \right)^2 - \frac{4}{9} + 1}\]
\[ = \frac{1}{6}\int\frac{\left( 6x - 4 \right) dx}{3 x^2 - 4x + 13} + \frac{1}{9}\int\frac{dx}{\left( x - \frac{2}{3} \right)^2 + \left( \frac{\sqrt{5}}{3} \right)^2}\]
\[ = \frac{1}{6} \text{ log } \left| 3 x^2 - 4x + 3 \right| + \frac{1}{9} \times \frac{3}{\sqrt{5}} \text{ tan }^{- 1} \left( \frac{x^{- \frac{2}{3}}}{\frac{\sqrt{5}}{3}} \right) + C\]
\[ = \frac{1}{6} \text{ log } \left| 3 x^2 - 4x + 3 \right| + \frac{1}{3\sqrt{5}} \text{ tan}^{- 1} \left( \frac{3 x - 2}{\sqrt{5}} \right) + C\]
\[ = \frac{1}{6} \text{ log }\left| 3 x^2 - 4x + 3 \right| + \frac{\sqrt{5}}{15} \text{ tan }^{- 1} \left( \frac{3x - 2}{\sqrt{5}} \right) + C\]