Evaluate the following Integrals :
`int ((x+2)dx)/((x^(2)+3x+3)sqrt(x+1))`

Answer

To evaluate the integral

\[ I = \int \frac{(x+2) \, dx}{(x^2 + 3x + 3) \sqrt{x+1}}, \]

we can follow these steps:

### Step 1: Simplify the Denominator
First, we can rewrite the denominator \(x^2 + 3x + 3\) in a more manageable form. Notice that:

\[ x^2 + 3x + 3 = (x + 1)^2 + 2. \]

Thus, we can express the integral as:

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

### Step 2: Substitution
Next, we will use the substitution \(u = \sqrt{x + 1}\). Then, we have:

\[ x = u^2 - 1 \quad \text{and} \quad dx = 2u \, du. \]

Now, substituting these into the integral gives:

\[ I = \int \frac{(u^2 - 1 + 2) \cdot 2u \, du}{((u^2)^2 - 2u^2 + 1 + 2) u} = \int \frac{(u^2 + 1) \cdot 2u \, du}{(u^4 - 2u^2 + 3) u}. \]

### Step 3: Simplify Further
This simplifies to:

\[ I = 2 \int \frac{(u^2 + 1) \, du}{u^4 - 2u^2 + 3}. \]

### Step 4: Break Down the Integral
We can break this integral into two parts:

\[ I = 2 \int \frac{u^2 \, du}{u^4 - 2u^2 + 3} + 2 \int \frac{du}{u^4 - 2u^2 + 3}. \]

### Step 5: Solve Each Integral
The first integral can be solved using partial fractions or a suitable substitution, while the second integral can be recognized as a standard form.

For the first integral, we can use the substitution \(v = u^2\), which gives us:

\[ I_1 = 2 \int \frac{du}{u^4 - 2u^2 + 3} = 2 \int \frac{1}{(u^2 - 1)^2 + 2} \, du. \]

This is a standard integral that can be solved using the arctangent function.

### Step 6: Combine Results
After evaluating both integrals, we will combine the results and substitute back \(u = \sqrt{x + 1}\) to express the final answer in terms of \(x\).

### Final Answer
The final expression for the integral will be:

\[ I = \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{x}{\sqrt{3} \sqrt{x + 1}} \right) + C, \]

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

---