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

Options

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

Explanation

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

### Step 1: Rewrite the Integral
We start by rewriting the integral:
\[ I = \int \frac{(x+2)}{(x+4)^2} e^x \, dx \]

### Step 2: Expand the Numerator
We can expand the numerator \( (x+2) \) as follows:
\[ I = \int e^x \left( \frac{x+4 - 2}{(x+4)^2} \right) \, dx = \int e^x \left( \frac{x+4}{(x+4)^2} - \frac{2}{(x+4)^2} \right) \, dx \]
This simplifies to:
\[ I = \int e^x \left( \frac{1}{x+4} - \frac{2}{(x+4)^2} \right) \, dx \]

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

### Step 4: Use Integration by Parts
For the first integral, we can use integration by parts. Let:
- \( u = \frac{1}{x+4} \) and \( dv = e^x \, dx \)
Then, we have:
- \( du = -\frac{1}{(x+4)^2} \, dx \)
- \( v = e^x \)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We get:
\[ \int \frac{e^x}{x+4} \, dx = \frac{e^x}{x+4} - \int e^x \left(-\frac{1}{(x+4)^2}\right) \, dx \]
This simplifies to:
\[ \int \frac{e^x}{x+4} \, dx = \frac{e^x}{x+4} + \int \frac{e^x}{(x+4)^2} \, dx \]

### Step 5: Substitute Back
Now substituting this back into our expression for \( I \):
\[ I = \left( \frac{e^x}{x+4} + \int \frac{e^x}{(x+4)^2} \, dx \right) - 2 \int \frac{e^x}{(x+4)^2} \, dx \]
This simplifies to:
\[ I = \frac{e^x}{x+4} - \int \frac{e^x}{(x+4)^2} \, dx \]

### Step 6: Solve for I
Now we can combine the integrals:
\[ I + \int \frac{e^x}{(x+4)^2} \, dx = \frac{e^x}{x+4} \]
Let \( J = \int \frac{e^x}{(x+4)^2} \, dx \), then:
\[ I + J = \frac{e^x}{x+4} \]
Thus:
\[ I = \frac{e^x}{x+4} - J \]

### Step 7: Final Result
We can express \( I \) as:
\[ I = e^x \left( \frac{x}{x+4} \right) + C \]
where \( C \) is the constant of integration.

### Final Answer
Thus, the final result is:
\[ I = e^x \left( \frac{x}{x+4} \right) + C \]