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

Explanation

To evaluate the integral

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

we will use a substitution method. Let's go through the steps:

### Step 1: Substitution
Let \( t = \log x \). Then, we have:

\[ dx = e^t dt. \]

### Step 2: Rewrite the Integral
Substituting \( t \) into the integral gives:

\[ \int \left( \log(t) + \frac{1}{t^2} \right) e^t dt. \]

### Step 3: Distributing \( e^t \)
We can distribute \( e^t \) across the terms in the integral:

\[ \int \log(t) e^t dt + \int \frac{1}{t^2} e^t dt. \]

### Step 4: Solve the First Integral
The first integral \( \int \log(t) e^t dt \) can be solved using integration by parts. Let:

- \( u = \log(t) \) → \( du = \frac{1}{t} dt \)
- \( dv = e^t dt \) → \( v = e^t \)

Using integration by parts:

\[ \int u \, dv = uv - \int v \, du, \]

we get:

\[ \int \log(t) e^t dt = \log(t) e^t - \int e^t \cdot \frac{1}{t} dt. \]

### Step 5: Solve the Second Integral
The second integral \( \int \frac{1}{t^2} e^t dt \) can also be solved using integration by parts. Let:

- \( u = \frac{1}{t} \) → \( du = -\frac{1}{t^2} dt \)
- \( dv = e^t dt \) → \( v = e^t \)

Using integration by parts again:

\[ \int \frac{1}{t^2} e^t dt = \frac{1}{t} e^t - \int e^t \cdot \left(-\frac{1}{t^2}\right) dt = \frac{1}{t} e^t + \int \frac{1}{t^2} e^t dt. \]

### Step 6: Combine Results
Now, combining both integrals, we have:

\[ \int \left( \log(t) + \frac{1}{t^2} \right) e^t dt = \log(t) e^t - \int \frac{1}{t} e^t dt + \frac{1}{t} e^t + C. \]

### Step 7: Substitute Back
Finally, substitute back \( t = \log x \):

\[ = \log(\log x) e^{\log x} - \int \frac{1}{\log x} e^{\log x} dx + \frac{1}{\log x} e^{\log x} + C. \]

Since \( e^{\log x} = x \), we can write:

\[ = \log(\log x) x - \int \frac{x}{\log x} dx + \frac{x}{\log x} + C. \]

### Final Answer
Thus, the evaluated integral is:

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