$$\int {{e^x}\left( {{{x - 1} \over {{x^2}}}} \right)dx = } $$

Area of the triangle formed by the lines $$y^2-9 x y+18 x^2=0$$ and $$y=9$$ is

The point on the curve $$y^2=2(x-3)$$ at which the normal is parallel to the line $$y-2 x+1=0$$ is

For two events $$\mathrm{A}$$ and $$\mathrm{B}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{5}{6}, \mathrm{P}(\mathrm{A})=\frac{1}{6}, \mathrm{P}(\mathrm{B})=\frac{2}{3}$$, then $$\mathrm{A}$$ and $$\mathrm{B}$$ are