Evaluate :
`int {log(logx)+(1)/((logx)^(2))}dx`

Answer

To evaluate the integral

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

we will use integration by parts.

### Step 1: Set up integration by parts
We can write the integral in the form suitable for integration by parts. Let:

- \( u = \log(\log x) \) (first function)
- \( dv = dx \) (second function)

Then, we differentiate and integrate to find \( du \) and \( v \):

- \( du = \frac{1}{\log x} \cdot \frac{1}{x} \, dx = \frac{1}{x \log x} \, dx \)
- \( v = x \)

### Step 2: Apply integration by parts
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we have:

\[ I = x \log(\log x) - \int x \cdot \frac{1}{x \log x} \, dx \]

This simplifies to:

\[ I = x \log(\log x) - \int \frac{1}{\log x} \, dx \]

### Step 3: Evaluate the remaining integral
Now we need to evaluate \( \int \frac{1}{\log x} \, dx \). We can again use integration by parts for this integral. Let:

- \( u = \frac{1}{\log x} \) (first function)
- \( dv = dx \) (second function)

Then we differentiate and integrate to find \( du \) and \( v \):

- \( du = -\frac{1}{(\log x)^2} \cdot \frac{1}{x} \, dx = -\frac{1}{x (\log x)^2} \, dx \)
- \( v = x \)

Applying integration by parts again:

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

This simplifies to:

\[ \int \frac{1}{\log x} \, dx = \frac{x}{\log x} + \int \frac{1}{(\log x)^2} \, dx \]

### Step 4: Substitute back
Now substituting back into our expression for \( I \):

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

This gives:

\[ I = x \log(\log x) - \frac{x}{\log x} - \int \frac{1}{(\log x)^2} \, dx \]

### Step 5: Final result
Thus, the final expression for the integral is:

\[ I = x \log(\log x) - \frac{x}{\log x} - \int \frac{1}{(\log x)^2} \, dx + C \]

where \( C \) is the constant of integration.