`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 \( \int \frac{x+3}{(x+4)^2} e^x \, dx \), we can follow these steps:

### Step 1: Rewrite the Numerator
We can rewrite the numerator \( x + 3 \) as \( (x + 4) - 1 \). This allows us to separate the integral into two parts:

\[ \int \frac{x+3}{(x+4)^2} e^x \, dx = \int \frac{(x+4) - 1}{(x+4)^2} e^x \, dx \]

### Step 2: Split the Integral
Now we can split the integral into two simpler integrals:

\[ \int \frac{(x+4)}{(x+4)^2} e^x \, dx - \int \frac{1}{(x+4)^2} e^x \, dx \]

This simplifies to:

\[ \int \frac{1}{x+4} e^x \, dx - \int \frac{1}{(x+4)^2} e^x \, dx \]

### Step 3: Solve the First Integral
Now we will solve the first integral \( \int \frac{1}{x+4} e^x \, dx \). We can use integration by parts here. Let:

- \( u = \frac{1}{x+4} \) so that \( du = -\frac{1}{(x+4)^2} dx \)
- \( dv = e^x dx \) so that \( v = e^x \)

Using integration by parts \( \int u \, dv = uv - \int v \, du \):

\[ \int \frac{1}{x+4} e^x \, dx = \frac{e^x}{x+4} - \int e^x \left(-\frac{1}{(x+4)^2}\right) dx \]

This gives us:

\[ \int \frac{1}{x+4} e^x \, dx = \frac{e^x}{x+4} + \int \frac{e^x}{(x+4)^2} \, dx \]

### Step 4: Substitute Back
Now we substitute this back into our expression:

\[ \left( \frac{e^x}{x+4} + \int \frac{e^x}{(x+4)^2} \, dx \right) - \int \frac{1}{(x+4)^2} e^x \, dx \]

Notice that the two integrals \( \int \frac{e^x}{(x+4)^2} \, dx \) cancel each other out:

\[ \int \frac{x+3}{(x+4)^2} e^x \, dx = \frac{e^x}{x+4} + C \]

### Final Answer
Thus, the integral \( \int \frac{x+3}{(x+4)^2} e^x \, dx \) is equal to:

\[ \frac{e^x}{x+4} + C \]