`intlogx(logx+2)dx=`

Options

A. `x(logx)^(2)+c`
B. `x(1+logx)^(2)+c`
C. `x[1+(logx)^(2)]+c`
D. `x(1+logx)+c`

Answer

To solve the integral \( \int \log x (\log x + 2) \, dx \), we can break it down into simpler parts. Let's go through the steps:

### Step 1: Expand the Integral
First, we can expand the integrand:
\[ \int \log x (\log x + 2) \, dx = \int (\log^2 x + 2 \log x) \, dx \]

### Step 2: Separate the Integral
Now, we can separate the integral into two parts:
\[ \int \log^2 x \, dx + 2 \int \log x \, dx \]

### Step 3: Solve \( \int \log x \, dx \)
To solve \( \int \log x \, dx \), we can use integration by parts. Let:
- \( u = \log x \) → \( du = \frac{1}{x} \, dx \)
- \( dv = dx \) → \( v = x \)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We get:
\[ \int \log x \, dx = x \log x - \int x \cdot \frac{1}{x} \, dx = x \log x - x + C \]

### Step 4: Solve \( \int \log^2 x \, dx \)
For \( \int \log^2 x \, dx \), we again use integration by parts:
Let:
- \( u = \log^2 x \) → \( du = 2 \log x \cdot \frac{1}{x} \, dx \)
- \( dv = dx \) → \( v = x \)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We get:
\[ \int \log^2 x \, dx = x \log^2 x - \int x \cdot 2 \log x \cdot \frac{1}{x} \, dx = x \log^2 x - 2 \int \log x \, dx \]

### Step 5: Substitute Back
Now, we substitute \( \int \log x \, dx \) back into our equation:
\[ \int \log^2 x \, dx = x \log^2 x - 2(x \log x - x) = x \log^2 x - 2x \log x + 2x \]

### Step 6: Combine the Results
Now we combine the results:
\[ \int \log x (\log x + 2) \, dx = \left( x \log^2 x - 2x \log x + 2x \right) + 2\left( x \log x - x \right) \]
\[ = x \log^2 x - 2x \log x + 2x + 2x \log x - 2x \]
\[ = x \log^2 x + 0 \]

### Final Answer
Thus, the final result is:
\[ \int \log x (\log x + 2) \, dx = x \log^2 x + C \]