$\int \frac{X + 1}{x (1 + x e^{x})} d x = 0$

L o g ∣ ∣ ∣ \begin{matrix} \frac{1 + x e^{x}}{x e^{x}} \end{matrix} ∣ ∣ ∣ + C

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L o g ∣ ∣ ∣ \begin{matrix} \frac{x e^{x}}{1 + x e^{x}} \end{matrix} ∣ ∣ ∣ + C

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L o g | \begin{matrix} x e^{x} (1 + x e^{x}) \end{matrix} | + C

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L o g (1 + x e^{x}) + C

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Solution

The correct option is B $L o g ∣ ∣ ∣ \begin{matrix} \frac{x e^{x}}{1 + x e^{x}} \end{matrix} ∣ ∣ ∣ + C$
$\int \frac{e^{x} (x + 1)}{x e^{x} (1 + x e^{x})} d x$
Put $x e^{x} = 1$

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y = x e^{x}

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

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