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

Let's re-verify the step-by-step differentiation to pinpoint the exact placement of the negative signs:

Step-by-Step Solution:

1. Identify the function and its derivative:

Let $f(x) = \cot x$

Then $f'(x) = -\csc^2 x \implies \csc^2 x = -f'(x)$

2. Substitute these into the integral:

$\int e^{-x} (\cot x + \csc^2 x) \, dx = \int e^{-x} [f(x) - f'(x)] \, dx$

3. Evaluate using Integration by Parts:

Let's expand $\int e^{-x} f(x) \, dx$ using integration by parts, taking $f(x)$ as the first function and $e^{-x}$ as the second:

$\int e^{-x} f(x) \, dx = f(x) \cdot (-e^{-x}) - \int f'(x) \cdot (-e^{-x}) \, dx$

$\int e^{-x} f(x) \, dx = -e^{-x} f(x) + \int e^{-x} f'(x) \, dx$

4. Combine everything back into the original expression:

$\int e^{-x} [f(x) - f'(x)] \, dx = \left[ -e^{-x} f(x) + \int e^{-x} f'(x) \, dx \right] - \int e^{-x} f'(x) \, dx$

The two identical integrals with opposite signs cancel out perfectly:

$= -e^{-x} f(x) + C$

5. Substitute $f(x) = \cot x$ back into the result:

$= -e^{-x} \cot x + C$