Question
Evaluate $ \int \frac{dx}{e^x - 1}$ (Note: C is a constant of integration.)
The correct answer is
$ \ln(1-e^{-x})+C$
Evaluating the Integral $ \int \frac{dx}{e^x - 1} $
To evaluate the integral $ \int \frac{dx}{e^x - 1} $, we can use algebraic manipulation and substitution.
Integral Manipulation
- Rewrite the integrand by multiplying the numerator and denominator by $ e^{-x} $: $ \int \frac{dx}{e^x - 1} = \int \frac{e^{-x} dx}{e^{-x}(e^x - 1)} = \int \frac{e^{-x} dx}{1 - e^{-x}} $
Substitution Method
- Let $ u = 1 - e^{-x} $.
- Calculate the differential $ du $: $ du = - e^{-x}(-1) dx = e^{-x} dx $
- Substitute $ u $ and $ du $ into the integral: $ \int \frac{du}{u} $
- Evaluate the resulting integral: $ \int \frac{du}{u} = \ln|u| + C $
- Substitute back $ u = 1 - e^{-x} $: $ \ln|1 - e^{-x}| + C $
- Considering the common form where $ 1 - e^{-x} > 0 $ (which implies $ e^{-x} < 1 $ or $ x > 0 $), the expression simplifies to: $ \ln(1 - e^{-x}) + C $
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