$\int \frac{f'(x)}{f(x) \log_e[f(x)]} dx$ is equal to

Given: $\int \frac{f'(x)}{f(x) \log_e[f(x)]} dx$

Whenever $f'(x)$ appears in the numerator and $f(x)$ in the denominator, substitution simplifies the integral.

Let $u = f(x)$

Then $du = f'(x)dx$

The integral becomes:

$\int \frac{1}{u \cdot \log_e(u)} du$

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Notice that $\log_e(u)$ appears in the denominator, and the derivative of $\log_e(u)$ is $\frac{1}{u}$.

Let $t = \log_e(u)$

Then $dt = \frac{1}{u} du$

The integral becomes:

$\int \frac{1}{t} dt$

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$\int \frac{1}{t} dt = \log_e(t) + C$

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Substituting back $t = \log_e(u)$:

$\log_e(t) + C = \log_e(\log_e(u)) + C$

Substituting back $u = f(x)$:

$\log_e(\log_e(u)) + C = \log_e(\log_e(f(x))) + C$

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Therefore, $\int \frac{f'(x)}{f(x) \log_e[f(x)]} dx = \log_e(\log_e[f(x)]) + C$