To evaluate the integral \( I = \int_0^1 x \log(1 + 2x) \, dx \), we will use integration by parts.
### Step 1: Choose \( u \) and \( dv \)
Let's set:
- \( u = \log(1 + 2x) \) (which is our logarithmic function)
- \( dv = x \, dx \) (which is our algebraic function)
### Step 2: Differentiate \( u \) and integrate \( dv \)
Now, we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[
du = \frac{d}{dx}(\log(1 + 2x)) \, dx = \frac{2}{1 + 2x} \, dx
\]
- Integrate \( dv \):
\[
v = \int x \, dx = \frac{x^2}{2}
\]
### Step 3: Apply the integration by parts formula
The integration by parts formula is:
\[
\int u \, dv = uv - \int v \, du
\]
Substituting our values:
\[
I = \left[ \log(1 + 2x) \cdot \frac{x^2}{2} \right]_0^1 - \int_0^1 \frac{x^2}{2} \cdot \frac{2}{1 + 2x} \, dx
\]
### Step 4: Evaluate the boundary term
Now calculate the boundary term:
\[
\left[ \log(1 + 2x) \cdot \frac{x^2}{2} \right]_0^1 = \left( \log(3) \cdot \frac{1^2}{2} \right) - \left( \log(1) \cdot \frac{0^2}{2} \right) = \frac{1}{2} \log(3) - 0 = \frac{1}{2} \log(3)
\]
### Step 5: Simplify the integral
Now, we need to simplify the integral:
\[
\int_0^1 \frac{x^2}{1 + 2x} \, dx
\]
We can perform polynomial long division on \( \frac{x^2}{1 + 2x} \):
\[
\frac{x^2}{1 + 2x} = \frac{x^2}{2x} - \frac{x}{2(1 + 2x)} = \frac{x}{2} - \frac{x}{2(1 + 2x)}
\]
Thus, we can rewrite the integral:
\[
\int_0^1 \frac{x^2}{1 + 2x} \, dx = \int_0^1 \left( \frac{x}{2} - \frac{x}{2(1 + 2x)} \right) \, dx
\]
### Step 6: Evaluate the two integrals
1. The first integral:
\[
\int_0^1 \frac{x}{2} \, dx = \frac{1}{2} \cdot \frac{x^2}{2} \Big|_0^1 = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}
\]
2. The second integral:
\[
\int_0^1 \frac{x}{2(1 + 2x)} \, dx
\]
Let \( t = 1 + 2x \), then \( dt = 2 \, dx \) or \( dx = \frac{dt}{2} \). When \( x = 0 \), \( t = 1 \) and when \( x = 1 \), \( t = 3 \):
\[
\int_1^3 \frac{(t - 1)}{2t} \cdot \frac{dt}{2} = \frac{1}{4} \int_1^3 \left( 1 - \frac{1}{t} \right) dt = \frac{1}{4} \left[ t - \log(t) \right]_1^3
\]
Evaluating gives:
\[
= \frac{1}{4} \left( 3 - \log(3) - (1 - 0) \right) = \frac{1}{4} \left( 2 - \log(3) \right)
\]
### Step 7: Combine results
Now we can combine everything:
\[
I = \frac{1}{2} \log(3) - \left( \frac{1}{4} - \frac{1}{4} \left( 2 - \log(3) \right) \right)
\]
This simplifies to:
\[
I = \frac{1}{2} \log(3) - \frac{1}{4} + \frac{1}{4} \left( 2 - \log(3) \right) = \frac{1}{2} \log(3) - \frac{1}{4} + \frac{1}{2} - \frac{1}{4} \log(3)
\]
Combining like terms gives:
\[
I = \frac{1}{4} \log(3) + \frac{1}{4} = \frac{1}{4} (\log(3) + 1)
\]
### Final Answer
Thus, the final result is:
\[
\int_0^1 x \log(1 + 2x) \, dx = \frac{1}{4} (\log(3) + 1)
\]
