`int e^(x)[((x+3))/((x+4)^(2))]dx`

Answer

To solve the integral \( \int e^{x} \frac{x+3}{(x+4)^{2}} \, dx \), we can use integration by parts and some algebraic manipulation. Here’s a step-by-step solution:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int e^{x} \frac{x+3}{(x+4)^{2}} \, dx \]

### Step 2: Simplify the Integrand
We can express the integrand in a more manageable form. Notice that:
\[ \frac{x+3}{(x+4)^{2}} = \frac{(x+4) - 1}{(x+4)^{2}} = \frac{1}{x+4} - \frac{1}{(x+4)^{2}} \]
Thus, we can rewrite the integral as:
\[ I = \int e^{x} \left( \frac{1}{x+4} - \frac{1}{(x+4)^{2}} \right) \, dx \]

### Step 3: Split the Integral
Now we can split the integral into two parts:
\[ I = \int e^{x} \frac{1}{x+4} \, dx - \int e^{x} \frac{1}{(x+4)^{2}} \, dx \]

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

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

### Step 5: Substitute Back
Now substituting back into our expression for \( I \):
\[ I = \left( e^{x} \frac{1}{x+4} + \int e^{x} \frac{1}{(x+4)^{2}} \, dx \right) - \int e^{x} \frac{1}{(x+4)^{2}} \, dx \]
The integrals of \( e^{x} \frac{1}{(x+4)^{2}} \) cancel out:
\[ I = e^{x} \frac{1}{x+4} \]

### Step 6: Final Result
Thus, the final result of the integral is:
\[ I = e^{x} \frac{1}{x+4} + C \]
where \( C \) is the constant of integration.

---