`int_(1)^(2)(dx)/(sqrtx-sqrt(x-1))`

Answer

To solve the integral

\[ I = \int_{1}^{2} \frac{dx}{\sqrt{x} - \sqrt{x-1}}, \]

we will follow these steps:

### Step 1: Rationalize the Denominator
To simplify the integral, we can rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator, which is \(\sqrt{x} + \sqrt{x-1}\):

\[ I = \int_{1}^{2} \frac{\sqrt{x} + \sqrt{x-1}}{(\sqrt{x} - \sqrt{x-1})(\sqrt{x} + \sqrt{x-1})} \, dx. \]

### Step 2: Simplify the Denominator
The denominator can be simplified using the difference of squares:

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

Thus, the integral simplifies to:

\[ I = \int_{1}^{2} (\sqrt{x} + \sqrt{x-1}) \, dx. \]

### Step 3: Split the Integral
Now we can split the integral into two separate integrals:

\[ I = \int_{1}^{2} \sqrt{x} \, dx + \int_{1}^{2} \sqrt{x-1} \, dx. \]

### Step 4: Evaluate Each Integral
1. **First Integral:**
\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2}. \]
Evaluating from 1 to 2:
\[ \left[ \frac{2}{3} x^{3/2} \right]_{1}^{2} = \frac{2}{3} (2^{3/2}) - \frac{2}{3} (1^{3/2}) = \frac{2}{3} (2\sqrt{2}) - \frac{2}{3} (1) = \frac{4\sqrt{2}}{3} - \frac{2}{3} = \frac{4\sqrt{2} - 2}{3}. \]

2. **Second Integral:**
Let \(u = x - 1\), then \(du = dx\) and when \(x = 1\), \(u = 0\) and when \(x = 2\), \(u = 1\):
\[ \int_{0}^{1} \sqrt{u} \, du = \frac{2}{3} u^{3/2} \bigg|_{0}^{1} = \frac{2}{3} (1) - 0 = \frac{2}{3}. \]

### Step 5: Combine the Results
Now we combine both integrals:

\[ I = \left( \frac{4\sqrt{2} - 2}{3} \right) + \frac{2}{3} = \frac{4\sqrt{2} - 2 + 2}{3} = \frac{4\sqrt{2}}{3}. \]

### Final Answer
Thus, the value of the integral is:

\[ I = \frac{4\sqrt{2}}{3}. \]

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