Question

\[\int\frac{\log \left( 1 - x \right)}{x^2} \text{ dx}\]

Answer 1

\[\text{ Let I }= \int\frac{\log \left( 1 - x \right)}{x^2}dx\]
\[ = \int \frac{1}{x^2}_{II} \log \left( 1_I - x \right) \text{ dx}\]
\[ = \text{ log }\left( 1 - x \right)\int x^{- 2} dx - \int\frac{- 1}{1 - x} \times \left( \frac{x^{- 2 + 1}}{- 2 + 1} \right) dx\]
\[ = \text{ log} \left( 1 - x \right) \left[ \frac{x^{- 2 + 1}}{- 2 + 1} \right] + \int\frac{- 1}{\left( 1 - x \right) x}dx\]
\[ = \text{ log} \left( 1 - x \right) \times \left( - \frac{1}{x} \right) + \int\frac{1}{x^2 - x}dx\]
\[ = - \frac{\text{ log} \left( 1 - x \right)}{x} + \int\frac{1}{x^2 - x + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}dx\]
\[ = - \frac{\text{ log }\left( 1 - x \right)}{x} + \int\frac{1}{\left( x - \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}dx\]
\[ = - \frac{\text{ log }\left( 1 - x \right)}{x} + \frac{1}{2 \times \frac{1}{2}} \text{ log} \left| \frac{x - \frac{1}{2} - \frac{1}{2}}{x - \frac{1}{2} + \frac{1}{2}} \right| + C\]
\[ = - \frac{\text{ log }\left( 1 - x \right)}{x} + \text{ log} \left| \frac{x - 1}{x} \right| + C\]
\[ = - \frac{\text{ log} \left( 1 - x \right)}{x} + \text{ log }\left| \left( x - 1 \right) \right| - \log x + C\]
\[ = - \frac{\text{ log} \left| 1 - x \right|}{x} + \text{ log }\left| 1 - x \right| - \text{ log }\left| x \right| + C\]
\[ = \left( 1 - \frac{1}{x} \right) \text{ log} \left| 1 - x \right| - \text{ log} \left| x \right| + C\]