Integrate the rational function: $\frac{1}{e^x - 1}$ (Hint: Put $e^x = t$)

Let $e^x = t$. Then $e^x dx = dt$,which implies $dx = \frac{dt}{t}$.
Substituting these into the integral,we get:
$\int \frac{1}{e^x - 1} dx = \int \frac{1}{t - 1} \cdot \frac{dt}{t} = \int \frac{1}{t(t - 1)} dt$.
Using partial fractions:
$\frac{1}{t(t - 1)} = \frac{A}{t} + \frac{B}{t - 1}$.
$1 = A(t - 1) + Bt$.
Setting $t = 0$,we get $A = -1$.
Setting $t = 1$,we get $B = 1$.
Thus,$\int \frac{1}{t(t - 1)} dt = \int \left( \frac{-1}{t} + \frac{1}{t - 1} \right) dt$.
$= -\log|t| + \log|t - 1| + C = \log\left| \frac{t - 1}{t} \right| + C$.
Substituting $t = e^x$ back:
$= \log\left| \frac{e^x - 1}{e^x} \right| + C$.