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

Options

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

Answer

To solve the integral \( I = \int \frac{x + \sqrt{x + 1}}{x + 2} \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( x + 1 = t^2 \). Then, we have:
\[ x = t^2 - 1 \]
Differentiating both sides gives:
\[ dx = 2t \, dt \]

### Step 2: Substitute in the Integral
Now substituting \( x \) and \( dx \) into the integral:
\[ I = \int \frac{(t^2 - 1) + t}{(t^2 - 1) + 2} \cdot 2t \, dt \]
This simplifies to:
\[ I = \int \frac{t^2 + t - 1}{t^2 + 1} \cdot 2t \, dt \]

### Step 3: Simplifying the Integral
Now we can rewrite the integral:
\[ I = 2 \int \frac{t^3 + t^2 - t}{t^2 + 1} \, dt \]
We can separate this into simpler integrals:
\[ I = 2 \left( \int t \, dt + \int \frac{-t}{t^2 + 1} \, dt \right) \]

### Step 4: Solve Each Integral
1. For \( \int t \, dt \):
\[ \int t \, dt = \frac{t^2}{2} \]

2. For \( \int \frac{-t}{t^2 + 1} \, dt \):
Let \( u = t^2 + 1 \), then \( du = 2t \, dt \) or \( dt = \frac{du}{2t} \):
\[ \int \frac{-t}{t^2 + 1} \, dt = -\frac{1}{2} \ln |t^2 + 1| \]

### Step 5: Combine Results
Combining the results, we have:
\[ I = 2 \left( \frac{t^2}{2} - \frac{1}{2} \ln |t^2 + 1| \right) + C \]
This simplifies to:
\[ I = t^2 - \ln |t^2 + 1| + C \]

### Step 6: Back Substitute
Recall that \( t^2 = x + 1 \):
\[ I = (x + 1) - \ln |(x + 1) + 1| + C \]
This gives us:
\[ I = x + 1 - \ln |x + 2| + C \]

### Step 7: Final Answer
Thus, the final result is:
\[ \int \frac{x + \sqrt{x + 1}}{x + 2} \, dx = x + 1 - \ln |x + 2| + C \]