$\int \frac{(x + 1)}{x (1 + x e^{x})} d x =$

log | \frac{e^{x}}{1 + x e^{x}} | + c

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- log | \frac{e^{x}}{1 + x e^{x}} | + c

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log | \frac{x e^{x}}{1 + x e^{x}} | + c

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- log | \frac{e^{x}}{1 - x e^{x}} | + c

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Solution

The correct option is C $log | \frac{x e^{x}}{1 + x e^{x}} | + c$
Multiply and divide the given expression by $e^{x}$ .
$\int \frac{e^{x} (x + 1)}{x e^{x} (1 + x e^{x})} d x$
Now let, $x e^{x} = t$
Differentiating both sides,
$e^{x} (x + 1) d x = d t$
Substituting the above obtained values into the given expression,
$\int \frac{d t}{t (t + 1)}$ = $\int \frac{t + 1 - t}{t (t + 1)} d t$
$= \int \frac{1}{t} d t - \int \frac{1}{t + 1} d t$
$= ln t - ln (t + 1) + c$
$= ln (\frac{t}{t + 1}) + c$
Putting back the value of $t$ ,
$= ln | (\frac{x e^{x}}{1 + x e^{x}}) | + c$

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