What is `int ln(x)^(2)` dx equal to?

Options

A. `2x.ln(x)-2x+c`
B. `(2)/(x)+c`
C. `2x.ln(x)+c`
D. `(2ln(x))/(x)-2x+c`

Answer

To solve the integral \( \int \ln(x)^2 \, dx \), we will use integration by parts. Here’s a step-by-step solution:

### Step 1: Choose \( u \) and \( dv \)
Let:
- \( u = \ln(x)^2 \)
- \( dv = dx \)

### Step 2: Differentiate \( u \) and integrate \( dv \)
Now we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = 2 \ln(x) \cdot \frac{1}{x} \, dx = \frac{2 \ln(x)}{x} \, dx \]
- Integrate \( dv \):
\[ v = \int dx = x \]

### Step 3: Apply the integration by parts formula
The integration by parts formula is:
\[ \int u \, dv = uv - \int v \, du \]
Substituting our values:
\[ \int \ln(x)^2 \, dx = x \ln(x)^2 - \int x \cdot \frac{2 \ln(x)}{x} \, dx \]
This simplifies to:
\[ \int \ln(x)^2 \, dx = x \ln(x)^2 - 2 \int \ln(x) \, dx \]

### Step 4: Solve \( \int \ln(x) \, dx \)
We need to compute \( \int \ln(x) \, dx \) using integration by parts again:
- Let \( u = \ln(x) \) and \( dv = dx \).
- Then, \( du = \frac{1}{x} \, dx \) and \( v = x \).

Applying integration by parts:
\[ \int \ln(x) \, dx = x \ln(x) - \int x \cdot \frac{1}{x} \, dx = x \ln(x) - \int dx = x \ln(x) - x + C \]

### Step 5: Substitute back into the original integral
Now substitute \( \int \ln(x) \, dx \) back into our expression:
\[ \int \ln(x)^2 \, dx = x \ln(x)^2 - 2 \left( x \ln(x) - x \right) + C \]
This simplifies to:
\[ \int \ln(x)^2 \, dx = x \ln(x)^2 - 2x \ln(x) + 2x + C \]

### Final Answer
Thus, the integral \( \int \ln(x)^2 \, dx \) is equal to:
\[ x \ln(x)^2 - 2x \ln(x) + 2x + C \]