Evaluate : `int(1)/(x^(2)). log x dx`

Options

A. `-(1/x)logx-(3/x)+C`
B. `-(1/x)log-(1/x)+C`
C. `(1/x)log-(1/x)+C`
D. `-(1/(2x))log-(1/x)+C`

Explanation

To evaluate the integral \( \int \frac{1}{x^2} \log x \, dx \), we will use integration by parts.

### Step-by-Step Solution:

1. **Identify the parts for integration by parts**:
We will use the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
Let:
\[ u = \log x \quad \text{and} \quad dv = \frac{1}{x^2} \, dx \]

2. **Differentiate \( u \) and integrate \( dv \)**:
Now we need to find \( du \) and \( v \):
\[ du = \frac{1}{x} \, dx \quad \text{and} \quad v = \int \frac{1}{x^2} \, dx = -\frac{1}{x} \]

3. **Apply the integration by parts formula**:
Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula:
\[ \int \frac{1}{x^2} \log x \, dx = uv - \int v \, du \]
This gives us:
\[ = \log x \left(-\frac{1}{x}\right) - \int \left(-\frac{1}{x}\right) \left(\frac{1}{x}\right) \, dx \]
Simplifying this:
\[ = -\frac{\log x}{x} + \int \frac{1}{x^2} \, dx \]

4. **Evaluate the remaining integral**:
Now we need to evaluate \( \int \frac{1}{x^2} \, dx \):
\[ \int \frac{1}{x^2} \, dx = -\frac{1}{x} \]

5. **Combine the results**:
Substitute back into our expression:
\[ = -\frac{\log x}{x} - \frac{1}{x} + C \]
where \( C \) is the constant of integration.

6. **Final result**:
Thus, the final answer is:
\[ \int \frac{1}{x^2} \log x \, dx = -\frac{\log x + 1}{x} + C \]