\( \int \frac { e ^ { - x } } { e ^ { - x } - 2 } \)

To solve the integral \( \int \frac { e ^ { - x } } { e ^ { - x } - 2 } \, dx \), let’s start by simplifying the integrand. Notice that we can rewrite the denominator. A convenient substitution is to let \( u = e^{-x} \), which transforms our integral into \( \int \frac{u}{u - 2} (-du) \). This simplifies to \( -\int \frac{u}{u - 2} \, du \). Performing polynomial long division on \( \frac{u}{u - 2} \), we can express it as \( 1 + \frac{2}{u - 2} \). Thus, our integral becomes: \[ -\int \left( 1 + \frac{2}{u - 2} \right) du = -\int 1 \, du - 2\int \frac{1}{u - 2} \, du \] This evaluates to: \[ -u - 2\ln|u - 2| + C \] Now back-substituting \( u = e^{-x} \), we have: \[ -e^{-x} - 2\ln|e^{-x} - 2| + C \] To recap, the solution to our integral is: \[ -e^{-x} - 2\ln|e^{-x} - 2| + C \] This is both elegant and provides insight into how exponential functions interact with logarithmic expressions!