Question

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

Answer 1

\[\int\frac{\left( 2x + 5 \right) dx}{x^2 - x - 2}\]
\[2x + 5 = A\frac{d}{dx}\left( x^2 - x - 2 \right) + B\]
\[2x + 5 = A \left( 2x - 1 \right) + B\]
\[2x + 5 = \left( 2 A \right) x + B - A\]

Comparing the Coefficients of like powers of x

\[2 A = 2\]
\[A = 1\]
\[B - A = 5\]
\[B - 1 = 5\]
\[B = 6\]

\[\therefore 2x + 5 = 1 \cdot \left( 2x - 1 \right) + 6\]
\[ \therefore \int\left( \frac{2x + 5}{x^2 - x - 2} \right)dx\]
\[ \Rightarrow \int\left( \frac{\left( 2x - 1 \right) + 6}{x^2 - x - 2} \right)dx\]
\[ \Rightarrow \int\left( \frac{2x - 1}{x^2 - x - 2} \right)dx + 6\int\frac{dx}{x^2 - x - 2}\]
\[ = I_1 + 6 I_2 \left( \text{ say }\right) . . . \left( 1 \right)\]
\[\text{ where }\]
\[ I_1 = \int\left( \frac{2x - 1}{x^2 - x - 2} \right)\text{ dx }I_2 = \int\frac{dx}{x^2 - x - 2}\]
\[ I_1 = \int\left( \frac{2x - 1}{x^2 - x - 2} \right)dx\]
\[\text{ let x}^2 - x - 2 = t\]
\[ \Rightarrow \left( 2x - 1 \right) dx = dt\]
\[ I_1 = \int\frac{dt}{t}\]
\[ I_1 = \text{ log }\left( t \right)\]
\[ I_1 = \text{ log }\left| x^2 - x - 2 \right| + C_1 . . . \left( 2 \right)\]
\[ I_2 = \int\frac{dx}{x^2 - x - 2}\]
\[ I_2 = \int\frac{dx}{x^2 - x + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2 - 2}\]
\[ I_2 = \int\frac{dx}{\left( x - \frac{1}{2} \right)^2 - \frac{1}{4} - 2}\]
\[ I_2 = \int\frac{dx}{\left( x - \frac{1}{2} \right)^2 - \left( \frac{3}{2} \right)^2}\]
\[ I_2 = \frac{1}{2 \times \frac{3}{2}} \text{ log }\left| \frac{x - \frac{1}{2} - \frac{3}{2}}{x - \frac{1}{2} + \frac{3}{2}} \right|\]
\[ I_2 = \frac{1}{3} \text{ log} \left| \frac{x - 2}{x + 1} \right| + C_2 . . . \left( 3 \right)\]
\[\int\frac{\left( 2x + 5 \right) dx}{x^2 - x - 2}\]
\[ = \text{ log} \left| x^2 - x - 2 \right| + \frac{6}{3} \text{ log} \left| \frac{x - 2}{x + 1} \right| + C_1 + C_2 \]
\[ = \text{ log }\left| x^2 - x - 2 \right| + 2 \text{ log }\left| \frac{x - 2}{x + 1} \right| + C \left( \text{ Where C }= C_1 + C_2 \right)\]