Find $\int \left[ \log(\log x) + \frac{1}{(\log x)^2} \right] dx$.

$x \log(\log x) - \frac{x}{\log x} + C$

The correct answer is Option (2) → $x \log(\log x) - \frac{x}{\log x} + C$

Let $I = \int \left[ \log(\log x) + \frac{1}{(\log x)^2} \right] dx$

$I = \int \log(\log x) dx + \int \frac{dx}{(\log x)^2}$

$I = I_1 + I_2 + C \text{ (Say)} \quad \{C = \text{Arbitrary constant}\}$

$\text{Where } I_1 = \int \log(\log x) dx \text{ \& } I_2 = \int \frac{dx}{(\log x)^2}$

Consider $I_1$:

$I_1 = \int \log(\log x) \cdot 1 dx$

Applying Integration by parts:

$I_1 = x \log(\log x) - \int x \cdot \frac{1}{\log x} \cdot \frac{1}{x} dx$

$I_1 = x \log(\log x) - \int \frac{dx}{\log x} \cdot 1$

Applying Integration by parts again to the second term:

$I_1 = x \log(\log x) - \left[ \frac{x}{\log x} - \int x \cdot \frac{-1}{(\log x)^2} \cdot \frac{1}{x} dx \right]$

$I_1 = x \log(\log x) - \frac{x}{\log x} - \int \frac{dx}{(\log x)^2}$

But $\int \frac{dx}{(\log x)^2} = I_2$, so:

$∴I = I_1 + I_2 + C = x \log(\log x) - \frac{x}{\log x} + C$