`int_(0)^(1)x log (1+2 x)dx`

Answer

To solve the integral \( I = \int_{0}^{1} x \log(1 + 2x) \, dx \), we will use the substitution method followed by integration by parts. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( t = \log(1 + 2x) \). Then, we can express \( x \) in terms of \( t \):
\[ 1 + 2x = e^t \implies 2x = e^t - 1 \implies x = \frac{e^t - 1}{2} \]

Next, we need to find \( dx \):
\[ dx = \frac{d}{dt}\left(\frac{e^t - 1}{2}\right) dt = \frac{1}{2} e^t dt \]

### Step 2: Change the limits of integration
When \( x = 0 \):
\[ t = \log(1 + 2 \cdot 0) = \log(1) = 0 \]

When \( x = 1 \):
\[ t = \log(1 + 2 \cdot 1) = \log(3) \]

So the limits change from \( x: 0 \to 1 \) to \( t: 0 \to \log(3) \).

### Step 3: Substitute into the integral
Now we can substitute \( x \), \( \log(1 + 2x) \), and \( dx \) into the integral:
\[ I = \int_{0}^{\log(3)} \left(\frac{e^t - 1}{2}\right) t \left(\frac{1}{2} e^t\right) dt \]
This simplifies to:
\[ I = \frac{1}{4} \int_{0}^{\log(3)} (e^t - 1) t e^t \, dt \]
\[ = \frac{1}{4} \int_{0}^{\log(3)} (t e^{2t} - t e^t) \, dt \]

### Step 4: Split the integral
We can split this into two separate integrals:
\[ I = \frac{1}{4} \left( \int_{0}^{\log(3)} t e^{2t} \, dt - \int_{0}^{\log(3)} t e^t \, dt \right) \]

### Step 5: Integration by parts
For the first integral \( I_1 = \int t e^{2t} \, dt \):
Let \( u = t \) and \( dv = e^{2t} dt \). Then \( du = dt \) and \( v = \frac{1}{2} e^{2t} \).
Using integration by parts:
\[ I_1 = \frac{1}{2} t e^{2t} - \frac{1}{2} \int e^{2t} dt = \frac{1}{2} t e^{2t} - \frac{1}{4} e^{2t} \]

Now evaluate from \( 0 \) to \( \log(3) \):
\[ I_1 = \left[ \frac{1}{2} t e^{2t} - \frac{1}{4} e^{2t} \right]_{0}^{\log(3)} \]
Calculating at the limits:
\[ = \left[ \frac{1}{2} \log(3) e^{2 \log(3)} - \frac{1}{4} e^{2 \log(3)} \right] - \left[ 0 - \frac{1}{4} \right] \]
\[ = \left[ \frac{1}{2} \log(3) \cdot 9 - \frac{9}{4} \right] + \frac{1}{4} \]
\[ = \frac{9}{2} \log(3) - \frac{9}{4} + \frac{1}{4} = \frac{9}{2} \log(3) - \frac{8}{4} = \frac{9}{2} \log(3) - 2 \]

For the second integral \( I_2 = \int t e^t \, dt \):
Let \( u = t \) and \( dv = e^t dt \). Then \( du = dt \) and \( v = e^t \).
Using integration by parts:
\[ I_2 = t e^t - e^t \]
Evaluate from \( 0 \) to \( \log(3) \):
\[ I_2 = \left[ t e^t - e^t \right]_{0}^{\log(3)} = \left[ \log(3) \cdot 3 - 3 \right] - [0 - 1] = 3 \log(3) - 3 + 1 = 3 \log(3) - 2 \]

### Step 6: Combine results
Now substitute \( I_1 \) and \( I_2 \) back into the expression for \( I \):
\[ I = \frac{1}{4} \left( \left( \frac{9}{2} \log(3) - 2 \right) - \left( 3 \log(3) - 2 \right) \right) \]
\[ = \frac{1}{4} \left( \frac{9}{2} \log(3) - 3 \log(3) \right) = \frac{1}{4} \left( \frac{9 - 6}{2} \log(3) \right) = \frac{3}{8} \log(3) \]

### Final Answer
Thus, the value of the integral is:
\[ I = \frac{3}{8} \log(3) \]