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 \( \int_{0}^{1} \frac{x^2 e^x}{x+1} \, dx \) in terms of \( A = \int_{0}^{1} \frac{e^x}{x+1} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral we want to evaluate:
\[ I = \int_{0}^{1} \frac{x^2 e^x}{x+1} \, dx \]
We can split \( x^2 \) as follows:
\[ I = \int_{0}^{1} \frac{x^2 e^x}{x+1} \, dx = \int_{0}^{1} \frac{(x^2 - 1 + 1) e^x}{x+1} \, dx = \int_{0}^{1} \frac{(x^2 - 1)e^x}{x+1} \, dx + \int_{0}^{1} \frac{e^x}{x+1} \, dx \]
This gives us:
\[ I = \int_{0}^{1} \frac{(x^2 - 1)e^x}{x+1} \, dx + A \]

### Step 2: Simplify the First Integral
Now, we need to evaluate the first integral:
\[ \int_{0}^{1} \frac{(x^2 - 1)e^x}{x+1} \, dx = \int_{0}^{1} \frac{(x-1)(x+1)e^x}{x+1} \, dx = \int_{0}^{1} (x-1)e^x \, dx \]
This simplifies to:
\[ \int_{0}^{1} (x-1)e^x \, dx = \int_{0}^{1} xe^x \, dx - \int_{0}^{1} e^x \, dx \]

### Step 3: Evaluate the Integrals
Now we evaluate the two integrals:
1. For \( \int_{0}^{1} e^x \, dx \):
\[ \int e^x \, dx = e^x \Big|_{0}^{1} = e - 1 \]

2. For \( \int_{0}^{1} xe^x \, dx \), we use integration by parts:
Let \( u = x \) and \( dv = e^x \, dx \).
Then \( du = dx \) and \( v = e^x \).
Thus,
\[ \int xe^x \, dx = x e^x \Big|_{0}^{1} - \int e^x \, dx = (1 \cdot e - 0) - (e - 1) = e - (e - 1) = 1 \]

### Step 4: Combine Results
Now we can combine the results:
\[ \int_{0}^{1} (x-1)e^x \, dx = 1 - (e - 1) = 2 - e \]
Thus, substituting back into our expression for \( I \):
\[ I = (2 - e) + A \]

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