Question

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

Answer 1

\[\int\left( \frac{x^2 + x + 1}{x^2 - x} \right)dx\]
\[\frac{x^2 + x + 1}{x^2 - x} = 1 + \frac{2x + 1}{x^2 - x}\]
\[ \therefore \int\left( \frac{x^2 + x + 1}{x^2 - x} \right)dx\]
\[ = \int\left( 1 + \frac{2x + 1}{x^2 - x} \right)dx\]
\[ = \int1 + \left( \frac{2x - 1 + 2}{x^2 - x} \right)dx\]
 ` =  ∫   dx + ∫   {(2x -1 ) dx }/ {x^2 -x } +  ∫  { 2  dx } / { x^2 - x + (1/2)^2 - (1/2) ^2} `
\[ = ∫ dx + \int\frac{\left( 2x - 1 \right) dx}{x^2 - x} + 2\int\frac{dx}{\left( x - \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}\]
\[ = x + \text{ log } \left| x^2 - x \right| + 2 \times \frac{1}{2 \times \frac{1}{2}}\text{ log }\left| \frac{x - \frac{1}{2} - \frac{1}{2}}{x - \frac{1}{2} + \frac{1}{2}} \right|\]
\[ = x + \text{ log } \left| x^2 - x \right| + 2 \text{  log } \left| \frac{x - 1}{x} \right| + C\]