First slide

Methods of integration

Question

If $\int \frac{d x}{1 + e^{x}} = x + f (x) + C,$ then $f (x)$ is equal to

Moderate

A

$\log (1 + e^{x})$
B

$\log \frac{1}{1 + e^{x}}$
C

$2 \log (1 + e^{x})$
D

$\frac{x + e^{x}}{1 + e^{x}}$

Solution

Let $1 + e^{x} = t \Rightarrow e^{x} = t - 1 \Rightarrow x = \log (t - 1)$
$so d x = \frac{1}{t - 1} d t$
Thus $I = \int \frac{d x}{1 + e^{x}} = \int \frac{d t}{t (t - 1)} = \int (\frac{1}{t - 1} - \frac{1}{t}) d t$
$\begin{array}{l} = \log (t - 1) - \log t + C \\ = \log e^{x} - \log (1 + e^{x}) + C \\ = x - \log (1 + e^{x}) + C \\ = x + \log \frac{1}{1 + e^{x}} + C \end{array}$

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