Question

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

Answer 1

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

Comparing the Coefficients of like powers of x

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

\[1 - 3x = - \frac{1}{2}\left( 6x + 4 \right) + 3\]
\[Now, \int\frac{\left( 1 - 3x \right) dx}{3 x^2 + 4x + 2}\]
\[ = \int\left( \frac{\frac{- 1}{2}\left( 6x + 4 \right) + 3}{3 x^2 + 4x + 2} \right)dx\]
\[ = - \frac{1}{2}\int\frac{\left( 6x + 4 \right) dx}{3 x^2 + 4x + 2} + 3\int\frac{dx}{3 x^2 + 4x + 2}\]
\[ = - \frac{1}{2} I_1 + 3 I_2 \left( \text{ say} \right) . . . \left( 1 \right)\]
\[\text{ where}\]
\[ I_1 = \int\frac{6x + 4}{3 x^2 + 4x + 2} \text{ and }I_2 = \int\frac{dx}{3 x^2 + 4x + 2}\]
\[ I_1 = \int\left( \frac{6x + 4}{3 x^2 + 4x + 2} \right)dx\]
\[\text{ let }3 x^2 + 4x + 2 = t\]
\[ \Rightarrow \left( 6x + 4 \right) dx = dt\]
\[ I_1 = \int\frac{dt}{t}\]
\[ = \text{ log }\left| t \right| + C_1 \]
\[ = \text{ log }\left| 3 x^2 + 4x + 2 \right| + C_1 . . . \left( 2 \right)\]
\[ I_2 = \int\frac{dx}{3 x^2 + 4x + 2}\]
\[ I_2 = \frac{1}{3}\int\frac{dx}{x^2 + \frac{4}{3}x + \frac{2}{3}}\]
\[ I_2 = \frac{1}{3}\int\frac{dx}{x^2 + \frac{4x}{x} + \left( \frac{2}{3} \right)^2 - \left( \frac{2}{3} \right)^2 + \frac{2}{3}}\]
\[ I_2 = \frac{1}{3}\int\frac{dx}{\left( x - \frac{2}{3} \right)^2 - \frac{4}{9} + \frac{2}{3}}\]
\[ I_2 = \frac{1}{3}\int\frac{dx}{\left( x + \frac{2}{3} \right)^2 + \frac{2}{9}}\]
\[ I_2 = \frac{1}{3}\int\frac{dx}{\left( x + \frac{2}{3} \right)^2 + \left( \frac{\sqrt{2}}{3} \right)^2}\]
\[ I_2 = \frac{1}{3} \times \frac{3}{\sqrt{2}} \tan^{- 1} \left( \frac{x + \frac{2}{3}}{\frac{\sqrt{2}}{3}} \right) + C_2 \]
\[ I_2 = \frac{1}{\sqrt{2}} \tan^{- 1} \left( \frac{3x + 2}{\sqrt{2}} \right) + C_2 . . . \left( 3 \right)\]
\[\text{ from } \left( 1 \right), \left( 2 \right)\text{ and }\left( 3 \right)\]
\[\int\frac{\left( 1 - 3x \right) dx}{3 x^2 + 4x + 2}\]
\[ = - \frac{1}{2} \text{ log }\left| 3 x^2 + 4x + 2 \right| + \frac{3 \times 1}{\sqrt{2}} \text{ tan}^{- 1} \left( \frac{3x + 2}{\sqrt{2}} \right) + C_1 + C_2 \]
\[ = - \frac{1}{2} \text{ log }\left| 3 x^2 + 4x + 2 \right| + \frac{3}{\sqrt{2}} \text{ tan}^{- 1} \left( \frac{3x + 2}{\sqrt{2}} \right) + C \left( \text{ Where C } = C_1 + C_2 \right)\]