If the integral $I = \int \left[log_e(log_e x)^2 + \frac{a}{log_e x}\right] dx = x log_e(log_e x)^2 + C$, where C is constant of integration. Then the value of $a$ is:

Given that $I = \int \left[\log_e(\log_e x)^2 + \frac{a}{\log_e x}\right] dx = x \log_e(\log_e x)^2 + C$

If $\int f(x) dx = F(x) + C$, then $\frac{d}{dx}[F(x)] = f(x)$

Differentiating both sides with respect to $x$:

$\log_e(\log_e x)^2 + \frac{a}{\log_e x} = \frac{d}{dx}\left[x \log_e(\log_e x)^2\right]$

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Using the product rule on the right side:

$\frac{d}{dx}\left[x \log_e(\log_e x)^2\right]$

$= (1) \cdot \log_e(\log_e x)^2 + x \cdot \frac{d}{dx}\left[\log_e(\log_e x)^2\right]$

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For $\frac{d}{dx}\left[\log_e(\log_e x)^2\right]$, using the chain rule:

$\frac{d}{dx}\left[\log_e(\log_e x)^2\right]$

$= \frac{1}{(\log_e x)^2} \cdot \frac{d}{dx}\left[(\log_e x)^2\right]$

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

$= \frac{2}{x \log_e x}$

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Substituting back:

$\frac{d}{dx}\left[x \log_e(\log_e x)^2\right]$

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

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

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Equating both sides:

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

Comparing coefficients:

$\frac{a}{\log_e x} = \frac{2}{\log_e x}$

Therefore, $a = 2$