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To find the area of the shaded region between the curves \( y = \sqrt{x + 2} \) and \( y = \frac{1}{x + 1} \) from \( x = 0 \) to \( x = 2 \), we'll follow these steps:

### Step 1: Set up the integral
The area \( A \) of the region between two curves \( y_1(x) \) and \( y_2(x) \) from \( x = a \) to \( x = b \) is given by:
\[
A = \int_{a}^{b} \left[ y_1(x) - y_2(x) \right] \, dx
\]
Here, \( y_1(x) = \sqrt{x + 2} \) (the upper curve) and \( y_2(x) = \frac{1}{x + 1} \) (the lower curve). The bounds are from \( x = 0 \) to \( x = 2 \).

### Step 2: Evaluate the integral
The integral becomes:
\[
A = \int_{0}^{2} \left( \sqrt{x + 2} - \frac{1}{x + 1} \right) \, dx
\]

Let's compute this integral step-by-step.

1. **Integral of** \( \sqrt{x + 2} \):
   - Use the substitution \( u = x + 2 \), so \( du = dx \).
   \[
   \int \sqrt{x + 2} \, dx = \int \sqrt{u} \, du = \frac{2}{3} u^{3/2} + C = \frac{2}{3} (x + 2)^{3/2}
   \]

2. **Integral of** \( \frac{1}{x + 1} \):
   - This is a standard integral:
   \[
   \int \frac{1}{x + 1} \, dx = \ln|x + 1| + C
   \]

So, the area is:
\[
A = \left[ \frac{2}{3} (x + 2)^{3/2} - \ln|x + 1| \right]_{0}^{2}
\]

### Step 3: Compute the definite integral
Substitute the bounds \( x = 2 \) and \( x = 0 \) into the expression:

1. At \( x = 2 \):
   \[
   \frac{2}{3} (2 + 2)^{3/2} - \ln(2 + 1) = \frac{2}{3} \cdot 4^{3/2} - \ln(3) = \frac{2}{3} \cdot 8 - \ln(3) = \frac{16}{3} - \ln(3)
   \]

2. At \( x = 0 \):
   \[
   \frac{2}{3} (0 + 2)^{3/2} - \ln(0 + 1) = \frac{2}{3} \cdot 2^{3/2} - \ln(1) = \frac{2 \cdot 2\sqrt{2}}{3} - 0 = \frac{4\sqrt{2}}{3}
   \]

So, the area is:
\[
A = \left( \frac{16}{3} - \ln(3) \right) - \frac{4\sqrt{2}}{3}
\]
\[
A = \frac{16}{3} - \frac{4\sqrt{2}}{3} - \ln(3)
\]

### Step 4: Simplify the final answer
Combine the terms to get the final area:
\[
A = \frac{16 - 4\sqrt{2}}{3} - \ln(3)
\]
This expression gives the exact area of the shaded region.

Would you like further details on any part of this process, or help with something else?

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### Related Questions:
1. How would the area change if the region was between \( x = 1 \) and \( x = 2 \) instead of \( x = 0 \) to \( x = 2 \)?
2. Can you determine where the two curves intersect?
3. What is the maximum distance between the two curves in the interval \( x = 0 \) to \( x = 2 \)?
4. How do you determine which curve is the upper one without plotting?
5. What happens to the area if the upper and lower curves are switched?

### Tip:
Always check the intersection points between curves to accurately set the limits of integration when finding areas between curves.

Learn how to calculate the area between the curves y = sqrt(x + 2) and y = 1/(x + 1) from x = 0 to x = 2 using integration. This detailed solution covers setting up and evaluating the integral step-by-step, suitable for advanced high school or college-level students.

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