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

(N/A) Let $I = \int \left[ \log(\log x) + \frac{1}{(\log x)^2} \right] dx$
$= \int \log(\log x) dx + \int \frac{1}{(\log x)^2} dx$
For the first integral,use integration by parts with $u = \log(\log x)$ and $dv = dx$. Then $du = \frac{1}{\log x} \cdot \frac{1}{x} dx$ and $v = x$.
$I = x \log(\log x) - \int x \cdot \frac{1}{x \log x} dx + \int \frac{1}{(\log x)^2} dx$
$I = x \log(\log x) - \int \frac{1}{\log x} dx + \int \frac{1}{(\log x)^2} dx$
Now,integrate $\int \frac{1}{\log x} dx$ by parts with $u = \frac{1}{\log x}$ and $dv = dx$. Then $du = -\frac{1}{(\log x)^2} \cdot \frac{1}{x} dx$ and $v = x$.
$\int \frac{1}{\log x} dx = \frac{x}{\log x} - \int x \left( -\frac{1}{x(\log x)^2} \right) dx = \frac{x}{\log x} + \int \frac{1}{(\log x)^2} dx$
Substituting this back into the expression for $I$:
$I = x \log(\log x) - \left( \frac{x}{\log x} + \int \frac{1}{(\log x)^2} dx \right) + \int \frac{1}{(\log x)^2} dx$
$I = x \log(\log x) - \frac{x}{\log x} - \int \frac{1}{(\log x)^2} dx + \int \frac{1}{(\log x)^2} dx$
$I = x \log(\log x) - \frac{x}{\log x} + C$