Evaluate the following definite integrals:
`int_(0)^(1)x log(1+x/2)dx`

Answer

To evaluate the definite integral \( I = \int_{0}^{1} x \log\left(1 + \frac{x}{2}\right) \, dx \), we can follow these steps:

### Step 1: Simplify the logarithm
Using the property of logarithms, we can rewrite the integral as:
\[ I = \int_{0}^{1} x \left( \log(2 + x) - \log(2) \right) \, dx \]
This gives us two separate integrals:
\[ I = \int_{0}^{1} x \log(2 + x) \, dx - \log(2) \int_{0}^{1} x \, dx \]

### Step 2: Evaluate the second integral
The second integral is straightforward:
\[ \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1}{2} \]
Thus, we have:
\[ I = \int_{0}^{1} x \log(2 + x) \, dx - \frac{1}{2} \log(2) \]

### Step 3: Evaluate the first integral using substitution
Now we focus on the integral \( \int_{0}^{1} x \log(2 + x) \, dx \). We can use the substitution \( u = 2 + x \), which implies \( du = dx \) and when \( x = 0 \), \( u = 2 \) and when \( x = 1 \), \( u = 3 \). Also, \( x = u - 2 \). Therefore, the integral becomes:
\[ \int_{2}^{3} (u - 2) \log(u) \, du \]
This can be split into two integrals:
\[ \int_{2}^{3} u \log(u) \, du - 2 \int_{2}^{3} \log(u) \, du \]

### Step 4: Evaluate \( \int_{2}^{3} u \log(u) \, du \) using integration by parts
Let \( v = \log(u) \) and \( dw = u \, du \). Then, \( dv = \frac{1}{u} \, du \) and \( w = \frac{u^2}{2} \). Using integration by parts:
\[ \int u \log(u) \, du = \frac{u^2}{2} \log(u) - \int \frac{u^2}{2} \cdot \frac{1}{u} \, du = \frac{u^2}{2} \log(u) - \frac{1}{2} \int u \, du \]
Calculating the integral:
\[ \int u \, du = \frac{u^2}{2} \]
Thus,
\[ \int u \log(u) \, du = \frac{u^2}{2} \log(u) - \frac{u^2}{4} \]
Now we evaluate this from \( 2 \) to \( 3 \):
\[ \left[ \frac{u^2}{2} \log(u) - \frac{u^2}{4} \right]_{2}^{3} \]
Calculating at the bounds:
- For \( u = 3 \):
\[ \frac{3^2}{2} \log(3) - \frac{3^2}{4} = \frac{9}{2} \log(3) - \frac{9}{4} \]
- For \( u = 2 \):
\[ \frac{2^2}{2} \log(2) - \frac{2^2}{4} = 2 \log(2) - 1 \]
Thus,
\[ \int_{2}^{3} u \log(u) \, du = \left( \frac{9}{2} \log(3) - \frac{9}{4} \right) - \left( 2 \log(2) - 1 \right) \]

### Step 5: Combine results
Now substituting back into our expression for \( I \):
\[ I = \left( \frac{9}{2} \log(3) - \frac{9}{4} - 2 \log(2) + 1 \right) - \frac{1}{2} \log(2) \]
Combining the logarithmic terms:
\[ I = \frac{9}{2} \log(3) - \frac{9}{4} - \frac{5}{2} \log(2) + 1 \]

### Final Result
Thus, the evaluated integral is:
\[ I = \frac{9}{2} \log(3) - \frac{5}{2} \log(2) + \frac{7}{4} \]