If `I= int_(0)^(1) (e^(t))/(1+t) dt`, then `p= int_(0)^(1) e^(t) log (1+t) dt`=

Options

A. I
B. 2I
C. `e log 2-I`
D. none

Answer

To solve the problem, we need to evaluate the integral \( p = \int_0^1 e^t \log(1+t) \, dt \) in terms of \( I = \int_0^1 \frac{e^t}{1+t} \, dt \).

### Step-by-Step Solution:

1. **Define the Integral**:
We start with the integral we want to evaluate:
\[ p = \int_0^1 e^t \log(1+t) \, dt \]

2. **Integration by Parts**:
We will use integration by parts, where we let:
- \( u = \log(1+t) \) (first function)
- \( dv = e^t \, dt \) (second function)

Then, we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = \frac{1}{1+t} \, dt \]
- Integrate \( dv \):
\[ v = e^t \]

3. **Apply Integration by Parts Formula**:
The integration by parts formula is:
\[ \int u \, dv = uv - \int v \, du \]
Applying this to our integral:
\[ p = \left[ \log(1+t) \cdot e^t \right]_0^1 - \int_0^1 e^t \cdot \frac{1}{1+t} \, dt \]

4. **Evaluate the Boundary Terms**:
Now we evaluate the boundary terms:
- At \( t = 1 \):
\[ \log(1+1) \cdot e^1 = \log(2) \cdot e \]
- At \( t = 0 \):
\[ \log(1+0) \cdot e^0 = \log(1) \cdot 1 = 0 \]
Therefore, the boundary terms give:
\[ \left[ \log(1+t) \cdot e^t \right]_0^1 = e \log(2) - 0 = e \log(2) \]

5. **Substitute Back into the Integral**:
Now substituting back into our expression for \( p \):
\[ p = e \log(2) - \int_0^1 \frac{e^t}{1+t} \, dt \]
Notice that the remaining integral is exactly \( I \):
\[ p = e \log(2) - I \]

### Final Expression:
Thus, we have:
\[ p = e \log(2) - I \]