∫ e^-x (cot x + cosec^2x)dx =

The correct answer is Option (2) → $-e^{-x}\cot x + c$, where c is an arbitrary constant.

Given integral: $\int e^{-x}(\cot x + \csc^2 x)\,dx$

Break it into two parts:

$= \int e^{-x} \cot x\,dx + \int e^{-x} \csc^2 x\,dx$

Use integration by parts on both terms.

For $\int e^{-x} \cot x\,dx$:

Let $u = \cot x$, $dv = e^{-x} dx$

$\Rightarrow du = -\csc^2 x dx$, $v = -e^{-x}$

$\int e^{-x} \cot x\,dx = -e^{-x} \cot x - \int (-e^{-x})(-\csc^2 x) dx$

$= -e^{-x} \cot x - \int e^{-x} \csc^2 x dx$

Add $\int e^{-x} \csc^2 x dx$ to both sides:

$\int e^{-x}(\cot x + \csc^2 x)\,dx = -e^{-x} \cot x + C$