$\int e^{x}\left\{\frac{1}{1-x}+\frac{1}{(x-1)^{2}}\right\} d x=?$

$\frac{e^{\pi}}{1-x}$

$\frac{x}{1-e^{x}}$

$\frac{e^{x}}{x-1}$

$\frac{e^{x}}{(x-1)^{2}}$