`int (x+3)/(x+4)^(2)e^(x)\ dx` is equal to

Options

A. `e^(x)((1)/(x+4))+C`
B. `e^(-x)((1)/(x+4))+C`
C. `e^(-x)((1)/(x-4))+C`
D. `e^(2x)((1)/(x-4))+C`

Explanation

To solve the integral \( I = \int \frac{x+3}{(x+4)^2} e^x \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We start by rewriting \( x + 3 \) as \( (x + 4) - 1 \):
\[ I = \int \frac{(x + 4) - 1}{(x + 4)^2} e^x \, dx \]
This simplifies to:
\[ I = \int \left( \frac{x + 4}{(x + 4)^2} - \frac{1}{(x + 4)^2} \right) e^x \, dx \]

### Step 2: Simplify the integral
Now, we can separate the integral into two parts:
\[ I = \int \frac{1}{x + 4} e^x \, dx - \int \frac{1}{(x + 4)^2} e^x \, dx \]

### Step 3: Identify the first integral
Let’s denote:
\[ f(x) = \frac{1}{x + 4} \]
Then, the derivative \( f'(x) \) is:
\[ f'(x) = -\frac{1}{(x + 4)^2} \]
Thus, we can rewrite the second integral:
\[ \int \frac{1}{(x + 4)^2} e^x \, dx = -\int f'(x) e^x \, dx \]

### Step 4: Apply the integration by parts formula
Using the formula for integration of the product of a function and its derivative:
\[ \int e^x f(x) + e^x f'(x) \, dx = e^x f(x) + C \]
We can combine the two integrals:
\[ I = \int e^x f(x) \, dx \]

### Step 5: Solve the integral
Now substituting back:
\[ I = e^x f(x) + C = e^x \cdot \frac{1}{x + 4} + C \]

### Final Answer
Thus, the integral evaluates to:
\[ I = \frac{e^x}{x + 4} + C \]

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