Question

\[\int\limits_0^2 \frac{1}{4 + x - x^2} dx\]

Answer 1

\[Let\ I = \int_0^2 \frac{1}{4 + x - x^2}\ d\ x\ . Then, \]
\[I = - \int_0^2 \frac{1}{x^2 - x - 4} d x\]
\[ \Rightarrow I = - \int_0^2 \frac{1}{\left( x^2 - x + \frac{1}{4} \right) - \frac{1}{4} - 4} d\ x\]
\[ = - \int_0^2 \frac{1}{\left( x - \frac{1}{2} \right)^2 - \frac{17}{4}} d x\]
\[ = - \int_0^2 \frac{1}{\left( x - \frac{1}{2} \right)^2 - \left( \frac{\sqrt{17}}{2} \right)^2} d\ x\]
\[ = \int_0^2 \frac{1}{- \left( \frac{2x - 1}{2} \right)^2 + \left( \frac{\sqrt{17}}{2} \right)^2} d\ x\]
\[ = \frac{1}{\sqrt{17}} \left[ \log \left( \frac{\sqrt{17} + 2x - 1}{\sqrt{17} - 2x + 1} \right) \right]_0^2 \]
\[ = \frac{1}{\sqrt{17}}\left\{ \log \frac{\sqrt{17} + 3}{\sqrt{17} - 3} - \log \frac{\sqrt{17} - 1}{\sqrt{17} + 1} \right\}\]
\[ = \frac{1}{\sqrt{17}}\left\{ \log \frac{26 + 6\sqrt{17}}{8} - \log \frac{18 - 2\sqrt{17}}{16} \right\}\]
\[ = \frac{1}{\sqrt{17}}\left\{ \log \frac{52 + 12\sqrt{17}}{18 - 2\sqrt{17}} \right\}\]
\[ = \frac{1}{\sqrt{17}}\left\{ \log \frac{52 + 12\sqrt{17}}{18 - 2\sqrt{17}} \times \frac{18 + 2\sqrt{17}}{18 + 2\sqrt{17}} \right\}\]
\[ \Rightarrow I = \frac{1}{\sqrt{17}} \log \frac{1344 + 320\sqrt{17}}{256}\]
\[ \Rightarrow I = \frac{1}{\sqrt{17}} \log \frac{21 + 5\sqrt{17}}{4}\]