`int_(-1)^(1)xln(1+e^(x))dx=`_____.

Answer

To solve the integral \( \int_{-1}^{1} x \ln(1 + e^{x}) \, dx \), we can use the property of definite integrals that states:

\[ \int_{-a}^{a} f(x) \, dx = \int_{0}^{a} f(x) \, dx + \int_{0}^{a} f(-x) \, dx \]

### Step-by-Step Solution:

1. **Apply the property of definite integrals**:
We can express the integral as:
\[ \int_{-1}^{1} x \ln(1 + e^{x}) \, dx = \int_{0}^{1} x \ln(1 + e^{x}) \, dx + \int_{0}^{1} (-x) \ln(1 + e^{-x}) \, dx \]

2. **Rewrite the second integral**:
The second integral can be rewritten as:
\[ \int_{0}^{1} -x \ln(1 + e^{-x}) \, dx = -\int_{0}^{1} x \ln(1 + e^{-x}) \, dx \]

3. **Combine the two integrals**:
Now we can combine the two integrals:
\[ \int_{-1}^{1} x \ln(1 + e^{x}) \, dx = \int_{0}^{1} x \ln(1 + e^{x}) \, dx - \int_{0}^{1} x \ln(1 + e^{-x}) \, dx \]

4. **Simplify the logarithm**:
We can simplify the logarithm:
\[ \ln(1 + e^{-x}) = \ln\left(\frac{1 + e^{x}}{e^{x}}\right) = \ln(1 + e^{x}) - x \]
Thus, we have:
\[ -\int_{0}^{1} x \ln(1 + e^{-x}) \, dx = -\int_{0}^{1} x \left(\ln(1 + e^{x}) - x\right) \, dx \]

5. **Combine the integrals**:
Now, substituting back, we get:
\[ \int_{-1}^{1} x \ln(1 + e^{x}) \, dx = \int_{0}^{1} x \ln(1 + e^{x}) \, dx + \int_{0}^{1} x^2 \, dx \]

6. **Evaluate the integral of \( x^2 \)**:
The integral \( \int_{0}^{1} x^2 \, dx \) can be calculated as:
\[ \int_{0}^{1} x^2 \, dx = \left[\frac{x^3}{3}\right]_{0}^{1} = \frac{1}{3} \]

7. **Final result**:
Therefore, the value of the original integral is:
\[ \int_{-1}^{1} x \ln(1 + e^{x}) \, dx = 0 + \frac{1}{3} = \frac{1}{3} \]

### Final Answer:
\[ \int_{-1}^{1} x \ln(1 + e^{x}) \, dx = \frac{1}{3} \]