`int(x+sqrt(x+1))/(x+2)` dx is equal to:

Options

A. `(x+1)-2sqrt(x+1)+2"ln"|x+2|=2tan^(-1)sqrt(x+1)+C`
B. `(x+1)+2sqrt(x+2)-2"ln"|x+2|-2tan^(-1)sqrt(x+2)+C`
C. `(x+1)+2sqrt(x+1)-2"ln"|x+2|-2tan^(-1)sqrt(x+1)+C`
D. `(x=1)+2sqrt(x+2)-2"ln"|x+1|+2tan^(-1)sqrt(x+2)+C`

Answer

To solve the integral \(\int \frac{x + \sqrt{x + 1}}{x + 2} \, dx\), we will use a substitution method. Let's go through the solution step by step.

### Step 1: Substitution
Let’s substitute \(t^2 = x + 1\). Then, we can express \(x\) in terms of \(t\):
\[ x = t^2 - 1 \]
Next, we differentiate \(t^2\) to find \(dx\):
\[ dx = 2t \, dt \]

### Step 2: Rewrite the Integral
Now we need to rewrite the integral in terms of \(t\):
\[ \sqrt{x + 1} = \sqrt{t^2} = t \]
And for the denominator:
\[ x + 2 = (t^2 - 1) + 2 = t^2 + 1 \]
Now substituting everything into the integral gives:
\[ \int \frac{(t^2 - 1) + t}{t^2 + 1} \cdot 2t \, dt \]

### Step 3: Simplify the Integral
This simplifies to:
\[ \int \frac{(t^2 - 1 + t)}{t^2 + 1} \cdot 2t \, dt = \int \frac{(t^3 + t^2 - t)}{t^2 + 1} \cdot 2 \, dt \]
Now we can separate the integral:
\[ = 2 \int \frac{t^3 + t^2 - t}{t^2 + 1} \, dt \]

### Step 4: Long Division
We can perform polynomial long division on \(\frac{t^3 + t^2 - t}{t^2 + 1}\):
1. Divide \(t^3\) by \(t^2\) to get \(t\).
2. Multiply \(t\) by \(t^2 + 1\) to get \(t^3 + t\).
3. Subtract: \((t^3 + t^2 - t) - (t^3 + t) = t^2 - 2t\).

So, we can rewrite the integral as:
\[ 2 \int \left(t + \frac{t^2 - 2t}{t^2 + 1}\right) dt \]

### Step 5: Integrate Each Term
Now we can integrate each term separately:
1. \(\int t \, dt = \frac{t^2}{2}\)
2. For \(\int \frac{t^2 - 2t}{t^2 + 1} \, dt\), we can split it into two parts:
- \(\int 1 \, dt = t\)
- \(\int \frac{-2t}{t^2 + 1} \, dt\) can be solved using substitution \(u = t^2 + 1\), leading to \(-\log(t^2 + 1)\).

Putting it all together, we have:
\[ 2 \left( \frac{t^2}{2} + t - \log(t^2 + 1) \right) + C = t^2 + 2t - 2\log(t^2 + 1) + C \]

### Step 6: Back Substitute
Now we substitute back \(t = \sqrt{x + 1}\):
\[ = (\sqrt{x + 1})^2 + 2\sqrt{x + 1} - 2\log((\sqrt{x + 1})^2 + 1) + C \]
This simplifies to:
\[ = (x + 1) + 2\sqrt{x + 1} - 2\log(x + 2) + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{x + \sqrt{x + 1}}{x + 2} \, dx = x + 1 + 2\sqrt{x + 1} - 2\log(x + 2) + C \]