`int(x+1)^(2)e^(x)dx` is equal to

Options

A. `xe^(x)+C`
B. `x^(2)e^(x)+C`
C. `(x+1)e^(x)+C`
D. `(x^(2)+1)e^(x)+C`

Explanation

To solve the integral \(\int (x+1)^2 e^x \, dx\), we can follow these steps:

### Step 1: Expand the integrand
First, we expand \((x+1)^2\):
\[ (x+1)^2 = x^2 + 2x + 1 \]
Thus, the integral becomes:
\[ \int (x^2 + 2x + 1) e^x \, dx \]

### Step 2: Split the integral
We can split the integral into three separate integrals:
\[ \int (x^2 + 2x + 1) e^x \, dx = \int x^2 e^x \, dx + 2\int x e^x \, dx + \int e^x \, dx \]

### Step 3: Solve each integral using integration by parts
We will use integration by parts for \(\int x^2 e^x \, dx\) and \(\int x e^x \, dx\).

#### For \(\int x^2 e^x \, dx\):
Let \(u = x^2\) and \(dv = e^x \, dx\). Then, \(du = 2x \, dx\) and \(v = e^x\).
Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
\[ \int x^2 e^x \, dx = x^2 e^x - \int e^x (2x) \, dx \]
This simplifies to:
\[ \int x^2 e^x \, dx = x^2 e^x - 2\int x e^x \, dx \]

#### For \(\int x e^x \, dx\):
Let \(u = x\) and \(dv = e^x \, dx\). Then, \(du = dx\) and \(v = e^x\).
Using integration by parts:
\[ \int x e^x \, dx = x e^x - \int e^x \, dx \]
\[ \int x e^x \, dx = x e^x - e^x \]
Thus, we have:
\[ \int x e^x \, dx = e^x (x - 1) \]

### Step 4: Substitute back into the integral
Now substituting back into our expression for \(\int x^2 e^x \, dx\):
\[ \int x^2 e^x \, dx = x^2 e^x - 2(e^x (x - 1)) \]
This simplifies to:
\[ \int x^2 e^x \, dx = x^2 e^x - 2e^x (x - 1) = x^2 e^x - 2xe^x + 2e^x \]
Combining terms, we get:
\[ \int x^2 e^x \, dx = e^x (x^2 - 2x + 2) \]

### Step 5: Combine all parts
Now we can combine all parts:
\[ \int (x^2 + 2x + 1) e^x \, dx = e^x (x^2 - 2x + 2) + 2(e^x (x - 1)) + e^x \]
This simplifies to:
\[ = e^x (x^2 - 2x + 2 + 2x - 2 + 1) = e^x (x^2 + 1) \]

### Final Answer
Thus, the final answer is:
\[ \int (x+1)^2 e^x \, dx = e^x (x^2 + 1) + C \]