Find `int[log(logx)+1/((logx)^2)]dx`

Answer

To solve the integral \( \int \left( \log(\log x) + \frac{1}{(\log x)^2} \right) dx \), we will use integration by parts. Let's break it down step by step.

### Step 1: Identify the parts for integration by parts
We will let:
- \( u = \log(\log x) \)
- \( dv = dx \)

Now we need 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 the integration by parts formula
The integration by parts formula is given by:
\[ \int u \, dv = uv - \int v \, du \]

Substituting our values into the formula:
\[ \int \log(\log x) \, dx = x \log(\log x) - \int x \cdot \frac{1}{x \log x} \, dx \]
This simplifies to:
\[ \int \log(\log x) \, dx = x \log(\log x) - \int \frac{1}{\log x} \, dx \]

### Step 3: Solve the remaining integral
Now we need to solve \( \int \frac{1}{\log x} \, dx \). We will again use integration by parts:
Let:
- \( u = \frac{1}{\log x} \)
- \( dv = dx \)

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

Now applying the integration by parts formula again:
\[ \int \frac{1}{\log x} \, dx = x \cdot \frac{1}{\log x} - \int x \left(-\frac{1}{x (\log x)^2}\right) \, dx \]
This simplifies to:
\[ \int \frac{1}{\log x} \, dx = \frac{x}{\log x} + \int \frac{1}{(\log x)^2} \, dx \]

### Step 4: Substitute back into the original integral
Now we substitute back into our expression for \( \int \log(\log x) \, dx \):
\[ \int \log(\log x) \, dx = x \log(\log x) - \left( \frac{x}{\log x} + \int \frac{1}{(\log x)^2} \, dx \right) \]
This gives us:
\[ \int \log(\log x) \, dx = x \log(\log x) - \frac{x}{\log x} - \int \frac{1}{(\log x)^2} \, dx \]

### Step 5: Combine everything
Now we can combine everything to find the final result:
\[ \int \left( \log(\log x) + \frac{1}{(\log x)^2} \right) dx = x \log(\log x) - \frac{x}{\log x} + C \]
where \( C \) is the constant of integration.

### Final Answer:
\[ \int \left( \log(\log x) + \frac{1}{(\log x)^2} \right) dx = x \log(\log x) - \frac{x}{\log x} + C \]