Evaluate the integral
`I = int_(0)^(1) (I n (1 + x))/( 1 + x) dx`

Answer

To evaluate the integral

\[ I = \int_{0}^{1} \frac{\ln(1 + x)}{1 + x} \, dx, \]

we will follow these steps:

### Step 1: Substitution
Let \( t = \ln(1 + x) \). Then, we differentiate both sides to find \( dx \) in terms of \( dt \).

\[ \frac{dt}{dx} = \frac{1}{1 + x} \implies dx = (1 + x) \, dt. \]

### Step 2: Change of Limits
Next, we need to change the limits of integration. When \( x = 0 \):

\[ t = \ln(1 + 0) = \ln(1) = 0. \]

When \( x = 1 \):

\[ t = \ln(1 + 1) = \ln(2). \]

Thus, the limits change from \( x = 0 \) to \( x = 1 \) into \( t = 0 \) to \( t = \ln(2) \).

### Step 3: Substitute in the Integral
Now we can substitute \( x \) in terms of \( t \). From \( t = \ln(1 + x) \), we have \( 1 + x = e^t \). Therefore,

\[ dx = e^t \, dt. \]

Substituting these into the integral gives:

\[ I = \int_{0}^{\ln(2)} \frac{t}{e^t} e^t \, dt = \int_{0}^{\ln(2)} t \, dt. \]

### Step 4: Evaluate the Integral
Now we can evaluate the integral:

\[ I = \int_{0}^{\ln(2)} t \, dt = \left[ \frac{t^2}{2} \right]_{0}^{\ln(2)} = \frac{(\ln(2))^2}{2} - \frac{0^2}{2} = \frac{(\ln(2))^2}{2}. \]

### Final Result
Thus, the value of the integral is:

\[ I = \frac{(\ln(2))^2}{2}. \]

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