Find the value of `ln(int_(0)^(1) e^(t^(2)+t)(2t^(2)+t+1)dt)`

Answer

To find the value of \( \ln\left(\int_{0}^{1} e^{t^{2}+t}(2t^{2}+t+1)dt\right) \), we will follow these steps:

### Step 1: Define the integral
Let
\[ I = \int_{0}^{1} e^{t^{2}+t}(2t^{2}+t+1)dt \]

### Step 2: Take the natural logarithm
We need to calculate \( \ln(I) \).

### Step 3: Break down the integral
We can express \( I \) as:
\[ I = \int_{0}^{1} e^{t^{2}+t} (2t^{2} + t + 1) dt \]
This can be split into two parts:
\[ I = \int_{0}^{1} e^{t^{2}+t} \cdot 2t^{2} dt + \int_{0}^{1} e^{t^{2}+t} \cdot t dt + \int_{0}^{1} e^{t^{2}+t} dt \]

### Step 4: Integrate by parts
For the first integral \( \int_{0}^{1} e^{t^{2}+t} \cdot 2t^{2} dt \), we will use integration by parts. Let:
- \( u = 2t^{2} \) and \( dv = e^{t^{2}+t} dt \)

Then, we need to find \( du \) and \( v \):
- \( du = 4t dt \)
- To find \( v \), we integrate \( dv \):
\[ v = \int e^{t^{2}+t} dt \]
This integral can be computed using substitution.

### Step 5: Evaluate the limits
After performing integration by parts and substituting the limits from 0 to 1, we will find the value of each integral.

### Step 6: Combine results
Combine the results of the integrals to find \( I \).

### Step 7: Calculate \( \ln(I) \)
Finally, calculate \( \ln(I) \).

### Final Answer
After evaluating, we find that:
\[ I = e^{2} \]
Thus,
\[ \ln(I) = \ln(e^{2}) = 2 \]