`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 \( I = \int \frac{x^2 e^x}{(x+2)^2} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral \( I \):
\[ I = \int \frac{x^2 e^x}{(x+2)^2} \, dx \]
We can add and subtract 4 in the numerator:
\[ I = \int \frac{x^2 - 4 + 4}{(x+2)^2} e^x \, dx \]
This simplifies to:
\[ I = \int \frac{(x^2 - 4) + 4}{(x+2)^2} e^x \, dx \]

### Step 2: Factor the Numerator
Notice that \( x^2 - 4 \) can be factored as:
\[ x^2 - 4 = (x-2)(x+2) \]
Thus, we can rewrite the integral:
\[ I = \int \frac{(x-2)(x+2) + 4}{(x+2)^2} e^x \, dx \]

### Step 3: Separate the Integral
Now we can separate the terms in the integral:
\[ I = \int \left( \frac{x-2}{x+2} + \frac{4}{(x+2)^2} \right) e^x \, dx \]

### Step 4: Simplify the Integral
This gives us two separate integrals:
\[ I = \int \frac{x-2}{x+2} e^x \, dx + \int \frac{4}{(x+2)^2} e^x \, dx \]

### Step 5: Solve the First Integral
For the first integral, we can use integration by parts. Let:
- \( u = \frac{x-2}{x+2} \) and \( dv = e^x \, dx \)

Then, we find \( du \) and \( v \):
\[ du = \frac{(x+2)(1) - (x-2)(1)}{(x+2)^2} \, dx = \frac{4}{(x+2)^2} \, dx \]
\[ v = e^x \]

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
This gives us:
\[ \int \frac{x-2}{x+2} e^x \, dx = \frac{x-2}{x+2} e^x - \int e^x \cdot \frac{4}{(x+2)^2} \, dx \]

### Step 6: Combine the Integrals
Now we can combine the results:
\[ I = \left( \frac{x-2}{x+2} e^x - \int \frac{4 e^x}{(x+2)^2} \, dx \right) + \int \frac{4 e^x}{(x+2)^2} \, dx \]
The integrals of \( \frac{4 e^x}{(x+2)^2} \) cancel out, leading to:
\[ I = \frac{x-2}{x+2} e^x + C \]

### Final Answer
Thus, the final answer is:
\[ I = \frac{x-2}{x+2} e^x + C \]