Evaluate the following integrals:
`int((x+1))/((x+2)^(2))e^(x)dx`

Answer

To evaluate the integral

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

we can start by rewriting the integrand.

### Step 1: Rewrite the integrand
We can express \(x + 1\) in terms of \(x + 2\):

\[ x + 1 = (x + 2) - 1. \]

Thus, we can rewrite the integral as:

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

### Step 2: Split the integral
Now we can split the integral into two parts:

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

Let’s denote these two integrals as \(I_1\) and \(I_2\):

\[ I_1 = \int \frac{1}{x + 2} e^x \, dx, \]
\[ I_2 = \int \frac{1}{(x + 2)^2} e^x \, dx. \]

### Step 3: Solve \(I_1\)
To solve \(I_1\), we can use integration by parts. Let:

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

Using integration by parts:

\[ I_1 = uv - \int v \, du = \frac{1}{x + 2} e^x - \int e^x \left(-\frac{1}{(x + 2)^2}\right) \, dx. \]

This gives us:

\[ I_1 = \frac{e^x}{x + 2} + I_2. \]

### Step 4: Substitute back into the expression for \(I\)
Now substituting \(I_1\) back into the expression for \(I\):

\[ I = \left(\frac{e^x}{x + 2} + I_2\right) - I_2 = \frac{e^x}{x + 2}. \]

### Step 5: Final result
Thus, the final result for the integral is:

\[ I = \frac{e^x}{x + 2} + C, \]

where \(C\) is the constant of integration.