First slide

Methods of integration

Question

The value of $\int \frac{(x + 1)}{x (1 + x e^{x})} d x$ is

Easy

A

$\log |1 + x e^{x}| + x + C$
B

$\frac{1}{2} \log |\frac{x (1 + x e^{x})}{x + 1}| + C$
C

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

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

Solution

Write the integrand as $\frac{1}{x} + 1 - \frac{(x + 1) e^{x}}{1 + x e^{x}}$
So the given integral is equal to
$\begin{array}{l} \log x + x - \log (1 + x e^{x}) + C \\ = \log x + \log e^{x} - \log (1 + x e^{x}) + C \\ = \log \frac{x e^{x}}{1 + x e^{x}} + C . \end{array}$

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