Let `A=int_(0)^(1)(e^(x))/(x+1)` dx then answer the following questions in terms of A.
Q. `int_(0)^(1)(x^(2)e^(x))/(x+1)dx` equals

Options

A. `A-e`
B. `e-2+A`
C. `2+A`
D. `2-e+A`

Explanation

To solve the integral \( I = \int_{0}^{1} \frac{x^2 e^x}{x+1} \, dx \) in terms of \( A = \int_{0}^{1} \frac{e^x}{x+1} \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral \( I \):
\[ I = \int_{0}^{1} \frac{x^2 e^x}{x+1} \, dx \]
We can express \( x^2 \) as \( (x+1)(x-1) + 1 \):
\[ I = \int_{0}^{1} \frac{(x+1)(x-1) + 1}{x+1} e^x \, dx \]
This simplifies to:
\[ I = \int_{0}^{1} (x-1)e^x \, dx + \int_{0}^{1} e^x \, dx \]

### Step 2: Split the Integral
Now we can split the integral into two parts:
\[ I = \int_{0}^{1} (x-1)e^x \, dx + \int_{0}^{1} e^x \, dx \]

### Step 3: Evaluate the First Integral
Next, we evaluate the first integral:
\[ \int_{0}^{1} (x-1)e^x \, dx = \int_{0}^{1} x e^x \, dx - \int_{0}^{1} e^x \, dx \]

### Step 4: Evaluate \( \int_{0}^{1} x e^x \, dx \)
To evaluate \( \int_{0}^{1} x e^x \, dx \), we can use integration by parts:
Let \( u = x \) and \( dv = e^x \, dx \).
Then, \( du = dx \) and \( v = e^x \).
Using integration by parts:
\[ \int x e^x \, dx = x e^x - \int e^x \, dx \]
Evaluating from 0 to 1:
\[ \int_{0}^{1} x e^x \, dx = \left[ x e^x \right]_{0}^{1} - \left[ e^x \right]_{0}^{1} \]
Calculating this:
\[ = (1 \cdot e^1 - 0) - (e^1 - e^0) = e - (e - 1) = 1 \]

### Step 5: Evaluate \( \int_{0}^{1} e^x \, dx \)
Now we evaluate \( \int_{0}^{1} e^x \, dx \):
\[ \int_{0}^{1} e^x \, dx = \left[ e^x \right]_{0}^{1} = e - 1 \]

### Step 6: Combine Results
Now substituting back into our expression for \( I \):
\[ I = (1 - (e - 1)) + (e - 1) = 1 - e + 1 + e - 1 = 2 - e \]

### Step 7: Relate to \( A \)
Since we have \( A = \int_{0}^{1} \frac{e^x}{x+1} \, dx \), we can express \( I \) in terms of \( A \):
\[ I = 2 - e + A \]

### Final Answer
Thus, the final result is:
\[ \int_{0}^{1} \frac{x^2 e^x}{x+1} \, dx = 2 - e + A \]