Evaluate: `int log x dx`

Answer

To evaluate the integral \( \int \log x \, dx \), we can use integration by parts. Let's go through the solution step by step.

### Step 1: Set up the integral
Let \( I = \int \log x \, dx \).

### Step 2: Choose \( u \) and \( dv \)
We will use integration by parts, which states:
\[ \int u \, dv = uv - \int v \, du \]
Here, we choose:
- \( u = \log x \) (which we will differentiate)
- \( dv = dx \) (which we will integrate)

### Step 3: Differentiate \( u \) and integrate \( dv \)
Now we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = \frac{1}{x} \, dx \]
- Integrate \( dv \):
\[ v = x \]

### Step 4: Apply integration by parts
Now we can substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula:
\[ I = \int \log x \, dx = x \log x - \int x \cdot \frac{1}{x} \, dx \]
This simplifies to:
\[ I = x \log x - \int 1 \, dx \]

### Step 5: Evaluate the remaining integral
Now we evaluate the remaining integral:
\[ \int 1 \, dx = x \]
Substituting this back into our equation gives:
\[ I = x \log x - x + C \]
where \( C \) is the constant of integration.

### Final Answer
Thus, the final result for the integral \( \int \log x \, dx \) is:
\[ \int \log x \, dx = x \log x - x + C \]

---