Evaluate: `intxlog(1+x)\ dx`

Answer

To evaluate the integral \( \int x \log(1+x) \, dx \), we will use the method of integration by parts. The formula for integration by parts is given by:

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

### Step 1: Choose \( u \) and \( dv \)
Let:
- \( u = \log(1+x) \) (which we will differentiate)
- \( dv = x \, dx \) (which we will integrate)

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

### Step 3: Apply the integration by parts formula
Substituting into the integration by parts formula:
\[ \int x \log(1+x) \, dx = \left( \log(1+x) \cdot \frac{x^2}{2} \right) - \int \left( \frac{x^2}{2} \cdot \frac{1}{1+x} \right) \, dx \]

### Step 4: Simplify the integral
Now we need to simplify the integral:
\[ \int \frac{x^2}{2(1+x)} \, dx = \frac{1}{2} \int \frac{x^2}{1+x} \, dx \]
We can simplify \( \frac{x^2}{1+x} \) by performing polynomial long division:
\[ \frac{x^2}{1+x} = x - 1 + \frac{1}{1+x} \]
Thus, we have:
\[ \int \frac{x^2}{1+x} \, dx = \int (x - 1 + \frac{1}{1+x}) \, dx \]

### Step 5: Integrate each term
Now we can integrate each term:
1. \( \int x \, dx = \frac{x^2}{2} \)
2. \( \int -1 \, dx = -x \)
3. \( \int \frac{1}{1+x} \, dx = \log(1+x) \)

Putting it all together:
\[ \int \frac{x^2}{1+x} \, dx = \frac{x^2}{2} - x + \log(1+x) \]

### Step 6: Substitute back into the equation
Now substituting back into our integration by parts result:
\[ \int x \log(1+x) \, dx = \frac{x^2}{2} \log(1+x) - \frac{1}{2} \left( \frac{x^2}{2} - x + \log(1+x) \right) \]

### Step 7: Simplify the expression
This simplifies to:
\[ = \frac{x^2}{2} \log(1+x) - \frac{1}{4} x^2 + \frac{1}{2} x - \frac{1}{2} \log(1+x) + C \]
Where \( C \) is the constant of integration.

### Final Result
Thus, the final result is:
\[ \int x \log(1+x) \, dx = \frac{x^2}{2} \log(1+x) - \frac{x^2}{4} + \frac{x}{2} - \frac{1}{2} \log(1+x) + C \]