Question

Evaluate the following integrals:

\[\int\frac{5x - 2}{1 + 2x + 3 x^2} \text{ dx }\]

Answer 1

\[\text{ Let I } = \int\frac{5x - 2}{1 + 2x + 3 x^2}dx\]
\[ = \int\frac{5x - 2}{3 x^2 + 2x + 1}dx\]
\[\text{We express}\ 5x - 2 = A\left( \frac{d}{d x}\left( 3 x^2 + 2x + 1 \right) \right) + B\]
\[5x - 2 = A(6x + 2) + B\]
\[\text{Equating the coefficients of x and constants, we get}\]
\[\text{ 5 = 6A and - 2 = 2A + B }\]
\[\text{ or A } = \frac{5}{6} \text{ and B }= - \frac{11}{3} \]
\[ \therefore I = \int\frac{\frac{5}{6}\left( 6x + 2 \right) - \frac{11}{3}}{3 x^2 + 2x + 1}dx\]
\[ = \frac{5}{6}\int\frac{\left( 6x + 2 \right)}{3 x^2 + 2x + 1}dx - \frac{11}{3}\int\frac{1}{3 x^2 + 2x + 1}dx\]
\[ = \frac{5}{6} I_1 - \frac{11}{3} I_2 . . . (1)\]
\[\text{ Now, } I_1 = \int\frac{\left( 6x + 2 \right)}{3 x^2 + 2x + 1}dx\]
\[ \text{ Let 3 x}^2 + 2x + 1 = t\]
\[ \text{On differentiating both sides, we get}\]
\[ \left( 6x + 2 \right)dx = dt\]
\[ \therefore I_1 = \int\frac{1}{t}dt\]
\[ = \text{ log }\left| t \right| + c_1 \]
\[ = \text{ log}\left| 3 x^2 + 2x + 1 \right| + c_1 . . . (2)\]
\[\text{ And,} I_2 = \int\frac{1}{3 x^2 + 2x + 1}dx\]
\[ = \frac{1}{3}\int\frac{1}{x^2 + \frac{2}{3}x + \frac{1}{3}}dx\]
\[ = \frac{1}{3}\int\frac{1}{x^2 + \frac{2}{3}x + \frac{1}{9} - \frac{1}{9} + \frac{1}{3}}dx\]
\[ = \frac{1}{3}\int\frac{1}{\left( x + \frac{1}{3} \right)^2 + \left( \frac{\sqrt{2}}{3} \right)^2}dx\]
\[ \text{ Let x } + \frac{1}{3} = t\]
\[ \text{On differentiating both sides, we get}\]
\[ dx = dt\]
\[ \therefore I_2 = \frac{1}{3}\int\frac{1}{\left( t \right)^2 + \left( \frac{\sqrt{2}}{3} \right)^2}dt\]
\[ = \frac{1}{3} \times \frac{1}{\frac{\sqrt{2}}{3}} \tan^{- 1} \frac{3t}{\sqrt{2}} + c_2 \]
\[ = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{3\left( x + \frac{1}{3} \right)}{\sqrt{2}} + c_2 \]
\[ = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{3x + 1}{\sqrt{2}} + c_2 . . . (3)\]
\[\text{ From (1), (2) and (3), we get}\]
\[ \therefore I = \frac{5}{6}\left( \log\left| 3 x^2 + 2x + 1 \right| + c_1 \right) - \frac{11}{3}\left( \frac{1}{\sqrt{2}} \tan^{- 1} \frac{3x + 1}{\sqrt{2}} + c_2 \right)\]
\[ = \frac{5}{6}\left( \log\left| 3 x^2 + 2x + 1 \right| \right) - \frac{11}{3}\left( \frac{1}{\sqrt{2}} \tan^{- 1} \frac{3x + 1}{\sqrt{2}} \right) + c\]
\[\text{ Hence }, \int\frac{5x - 2}{1 + 2x + 3 x^2}dx = \frac{5}{6}\left( \log\left| 3 x^2 + 2x + 1 \right| \right) - \frac{11}{3}\left( \frac{1}{\sqrt{2}} \tan^{- 1} \frac{3x + 1}{\sqrt{2}} \right) + c\]