$\int \frac{\log_e x}{(1 + \log_e x)^2} dx$ is equal to

When faced with a complicated integral, checking if differentiating the answer options works faster can save time and reduce errors.

Consider the option: $\frac{x}{1 + \log_e x} + C$

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To find: $\frac{d}{dx}\left(\frac{x}{1 + \log_e x}\right)$

Let $u = x$ and $v = 1 + \log_e x$

Then $u' = 1$ and $v' = \frac{1}{x}$

$\frac{d}{dx}\left(\frac{x}{1 + \log_e x}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}$

$= \frac{(1 + \log_e x) \cdot 1 - x \cdot \frac{1}{x}}{(1 + \log_e x)^2}$

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$= \frac{(1 + \log_e x) - 1}{(1 + \log_e x)^2}$

$= \frac{\log_e x}{(1 + \log_e x)^2}$

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This matches the original integrand.

Therefore, $\int \frac{\log_e x}{(1 + \log_e x)^2} dx = \frac{x}{1 + \log_e x} + C$