First slide

Methods of integration

Question

If the primitive of $\frac{1}{{(e^{x} - 1)}^{2}}$ is
$f (x) - \log | g (x) | + C$ then

Moderate

A

$dom f = R$
B

$g (x) = 1 - e^{x}$
C

$dom f = R ~ {0}$
D

$f (x) = 1 - e^{- x}$

Solution

$\int \frac{1}{{(e^{x} - 1)}^{2}} d x = \int \frac{e^{x} d x}{e^{x} {(e^{x} - 1)}^{2}} = \int \frac{d t}{t (t - 1)^{2}}$
$(t = e^{x})$
$\begin{aligned} = \int \frac{1}{t - 1} [\frac{1}{t - 1} - \frac{1}{t}] d t \\ = \int (\frac{1}{(t - 1)^{2}} - \frac{1}{t - 1} + \frac{1}{t}) d t \\ = \frac{1}{1 - t} - \log | t - 1 | + \log | t | + C \\ = {(1 - e^{x})}^{- 1} - \log |1 - e^{- x}| + C \end{aligned}$
Hence $f (x) = {(1 - e^{x})}^{- 1}$ and $g (x) = 1 - e^{- x}$
The domain of $f = R ~ {0}$

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