Evaluate: `int (dx)/(sqrt(x+3)-sqrt(x+2))`.

Answer

To evaluate the integral

\[ \int \frac{dx}{\sqrt{x+3} - \sqrt{x+2}}, \]

we will follow a series of steps to simplify and compute the integral.

### Step 1: Rationalize the Denominator

We start by rationalizing the denominator. We multiply the numerator and denominator by the conjugate of the denominator:

\[ \int \frac{dx}{\sqrt{x+3} - \sqrt{x+2}} \cdot \frac{\sqrt{x+3} + \sqrt{x+2}}{\sqrt{x+3} + \sqrt{x+2}}. \]

This gives us:

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

### Step 2: Simplify the Denominator

Using the difference of squares, we can simplify the denominator:

\[ (\sqrt{x+3})^2 - (\sqrt{x+2})^2 = (x+3) - (x+2) = 1. \]

Thus, the integral simplifies to:

\[ \int (\sqrt{x+3} + \sqrt{x+2}) \, dx. \]

### Step 3: Separate the Integral

Now we can separate the integral into two parts:

\[ \int \sqrt{x+3} \, dx + \int \sqrt{x+2} \, dx. \]

### Step 4: Compute Each Integral

For the first integral, let \( u = x + 3 \), then \( du = dx \):

\[ \int \sqrt{x+3} \, dx = \int \sqrt{u} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2} = \frac{2}{3} (x+3)^{3/2}. \]

For the second integral, let \( v = x + 2 \), then \( dv = dx \):

\[ \int \sqrt{x+2} \, dx = \int \sqrt{v} \, dv = \frac{v^{3/2}}{3/2} = \frac{2}{3} v^{3/2} = \frac{2}{3} (x+2)^{3/2}. \]

### Step 5: Combine Results

Combining both results, we have:

\[ \int \frac{dx}{\sqrt{x+3} - \sqrt{x+2}} = \frac{2}{3} (x+3)^{3/2} + \frac{2}{3} (x+2)^{3/2} + C, \]

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

### Final Result

Thus, the final result of the integral is:

\[ \int \frac{dx}{\sqrt{x+3} - \sqrt{x+2}} = \frac{2}{3} (x+3)^{3/2} + \frac{2}{3} (x+2)^{3/2} + C. \]

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