Question

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

Answer 1

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

Comparing the Coefficients of like powers of x we get

\[2A = 1 \Rightarrow A = \frac{1}{2}\]
\[3A + B = 0\]
\[\frac{3}{2} + B = 0\]
\[B = - \frac{3}{2}\]
\[x = \frac{1}{2} \left( 2x + 3 \right) - \frac{3}{2}\]

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