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

Options

A. `((x-1)/(x+2))e^(x)+c`
B. `((x-2)/(x+2))e^(x-1)+c`
C. `((x-2)/(x+2))e^(x)+c`
D. none of these

Answer

To solve the integral \( \int \frac{x^2 e^x}{(x+2)^2} \, dx \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Rewrite the Integral:**
We start with the integral:
\[ I = \int \frac{x^2 e^x}{(x+2)^2} \, dx \]
We can manipulate the numerator by adding and subtracting 4:
\[ I = \int \frac{x^2 - 4 + 4}{(x+2)^2} e^x \, dx \]
This can be rewritten as:
\[ I = \int \frac{(x^2 - 4) + 4}{(x+2)^2} e^x \, dx \]

2. **Factor the Quadratic:**
Notice that \( x^2 - 4 \) can be factored:
\[ x^2 - 4 = (x+2)(x-2) \]
Thus, we can rewrite the integral as:
\[ I = \int \left( \frac{(x+2)(x-2)}{(x+2)^2} + \frac{4}{(x+2)^2} \right) e^x \, dx \]
Simplifying gives:
\[ I = \int \left( \frac{x-2}{x+2} + \frac{4}{(x+2)^2} \right) e^x \, dx \]

3. **Split the Integral:**
Now we can split the integral into two parts:
\[ I = \int \frac{x-2}{x+2} e^x \, dx + \int \frac{4}{(x+2)^2} e^x \, dx \]

4. **Use Integration by Parts:**
For the first integral \( \int \frac{x-2}{x+2} e^x \, dx \), we can use integration by parts. Let:
\[ u = \frac{x-2}{x+2}, \quad dv = e^x \, dx \]
Then, we differentiate and integrate:
\[ du = \left( \frac{(x+2)(1) - (x-2)(1)}{(x+2)^2} \right) \, dx = \frac{4}{(x+2)^2} \, dx, \quad v = e^x \]
Applying integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
This gives:
\[ \int \frac{x-2}{x+2} e^x \, dx = \frac{x-2}{x+2} e^x - \int e^x \cdot \frac{4}{(x+2)^2} \, dx \]

5. **Combine the Integrals:**
Now substituting back into our expression for \( I \):
\[ I = \left( \frac{x-2}{x+2} e^x - \int e^x \cdot \frac{4}{(x+2)^2} \, dx \right) + \int \frac{4}{(x+2)^2} e^x \, dx \]
The two integrals involving \( \frac{4}{(x+2)^2} e^x \) cancel out:
\[ I = \frac{x-2}{x+2} e^x + C \]

### Final Answer:
Thus, the integral evaluates to:
\[ \int \frac{x^2 e^x}{(x+2)^2} \, dx = e^x \left( \frac{x-2}{x+2} \right) + C \]