Evaluate :
`int[(1)/(logx)-(1)/((log x)^(2))]dx`

Answer

To evaluate the integral

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

we will follow these steps:

### Step 1: Substitution
Let \( t = \log x \). Then, we differentiate both sides with respect to \( x \):

\[ \frac{dt}{dx} = \frac{1}{x} \implies dx = x \, dt. \]

Since \( x = e^t \), we can substitute \( dx \) as follows:

\[ dx = e^t \, dt. \]

### Step 2: Rewrite the Integral
Now, substituting \( t \) and \( dx \) into the integral, we have:

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

### Step 3: Distribute \( e^t \)
Distributing \( e^t \) gives us:

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

### Step 4: Split the Integral
We can split this into two separate integrals:

\[ \int \frac{e^t}{t} \, dt - \int \frac{e^t}{t^2} \, dt. \]

### Step 5: Evaluate the Integrals
The first integral, \( \int \frac{e^t}{t} \, dt \), does not have a simple elementary form, but we can denote it as \( \text{Ei}(t) \) (the Exponential Integral function). The second integral can be evaluated using integration by parts. Let:

- \( u = \frac{1}{t^2} \) and \( dv = e^t \, dt \).
- Then, \( du = -\frac{2}{t^3} \, dt \) and \( v = e^t \).

Using integration by parts:

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

we get:

\[ \int \frac{e^t}{t^2} \, dt = \frac{e^t}{t^2} - \int e^t \left(-\frac{2}{t^3}\right) dt. \]

This integral can be complex, but for simplicity, we can denote it as \( I(t) \).

### Step 6: Combine Results
Thus, the integral becomes:

\[ \text{Ei}(t) - \left( \frac{e^t}{t^2} + I(t) \right) + C, \]

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

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

\[ \text{Ei}(\log x) - \left( \frac{x}{(\log x)^2} + I(\log x) \right) + C. \]

### Final Answer
The final answer is:

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