` int(x+2)/((x^(2)+3x+3)sqrt(x+1))dx " is equal to"`

Options

A. `(1)/(sqrt3)"tan"^(-1)((x)/(sqrt(3(x+1))))`
B. `(2)/(sqrt3)"tan"^(-1)((x)/(sqrt(3(x+1))))`
C. `(2)/(sqrt3)"tan"^(-1)((x)/(sqrt(x+1)))`
D. non of these

Explanation

To solve the integral

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

we will use a substitution method.

### Step 1: Substitution
Let \( t^2 = x + 1 \). Then, we have:

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

### Step 2: Substitute in the integral
Substituting \( x \) and \( dx \) into the integral:

\[ \int \frac{(t^2 - 1) + 2}{((t^2 - 1)^2 + 3(t^2 - 1) + 3) \sqrt{t^2}} \cdot 2t \, dt. \]

This simplifies to:

\[ \int \frac{t^2 + 1}{((t^2 - 1)^2 + 3(t^2 - 1) + 3) t} \cdot 2t \, dt. \]

### Step 3: Simplify the denominator
Now, we need to simplify the denominator:

\[ (t^2 - 1)^2 + 3(t^2 - 1) + 3 = t^4 - 2t^2 + 1 + 3t^2 - 3 + 3 = t^4 + t^2 + 1. \]

So, the integral becomes:

\[ 2 \int \frac{t^2 + 1}{t^4 + t^2 + 1} \, dt. \]

### Step 4: Split the integral
We can split the integral:

\[ 2 \int \frac{t^2}{t^4 + t^2 + 1} \, dt + 2 \int \frac{1}{t^4 + t^2 + 1} \, dt. \]

### Step 5: Further simplification
For the first integral, we can use the substitution \( u = t^2 \), which gives \( du = 2t \, dt \) or \( dt = \frac{du}{2\sqrt{u}} \):

\[ \int \frac{u}{u^2 + u + 1} \frac{du}{2\sqrt{u}}. \]

And for the second integral, we can use a trigonometric substitution or partial fraction decomposition.

### Step 6: Solve the integrals
After working through the integrals, we will arrive at:

\[ \int \frac{1}{t^4 + t^2 + 1} \, dt = \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{t - \frac{1}{t}}{\sqrt{3}} \right) + C. \]

### Step 7: Back substitute
Finally, we substitute back \( t = \sqrt{x + 1} \):

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

### Final Answer
Thus, the integral evaluates to:

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