`int _(2) ^(3) (dx )/( (x-1) sqrt(x ^(2) - 2x ))`

Answer

To solve the integral

\[ I = \int_{2}^{3} \frac{dx}{(x-1) \sqrt{x^2 - 2x}} \]

we will use a substitution method. Let's go through the steps:

### Step 1: Substitution
We start by making the substitution \( t^2 = x^2 - 2x \). This implies:

\[ x^2 - 2x = t^2 \implies x^2 - 2x + 1 = t^2 + 1 \implies (x-1)^2 = t^2 + 1 \]

Now, differentiating \( t^2 = x^2 - 2x \):

\[ 2t \frac{dt}{dx} = 2x - 2 \implies \frac{dt}{dx} = \frac{x-1}{t} \implies dx = \frac{t}{x-1} dt \]

### Step 2: Rewrite the Integral
Substituting \( dx \) into the integral, we have:

\[ I = \int \frac{t \, dt}{(x-1) \sqrt{x^2 - 2x}} \]

Since \( \sqrt{x^2 - 2x} = t \), we can rewrite the integral as:

\[ I = \int \frac{t \, dt}{(x-1) t} = \int \frac{dt}{x-1} \]

### Step 3: Express \( x-1 \) in terms of \( t \)
From our earlier substitution, we have:

\[ (x-1)^2 = t^2 + 1 \implies x-1 = \sqrt{t^2 + 1} \]

Thus, the integral becomes:

\[ I = \int \frac{dt}{\sqrt{t^2 + 1}} \]

### Step 4: Integrate
The integral \( \int \frac{dt}{\sqrt{t^2 + 1}} \) is known to be:

\[ \int \frac{dt}{\sqrt{t^2 + 1}} = \ln |t + \sqrt{t^2 + 1}| + C \]

### Step 5: Resubstitute and Evaluate Limits
Now we need to change the limits back to the original variable \( x \). When \( x = 2 \):

\[ t^2 = 2^2 - 2 \cdot 2 = 0 \implies t = 0 \]

When \( x = 3 \):

\[ t^2 = 3^2 - 2 \cdot 3 = 3 \implies t = \sqrt{3} \]

Thus, our limits change from \( t = 0 \) to \( t = \sqrt{3} \).

Now we can evaluate:

\[ I = \left[ \ln |t + \sqrt{t^2 + 1}| \right]_{0}^{\sqrt{3}} = \ln(\sqrt{3} + 2) - \ln(0 + 1) = \ln(\sqrt{3} + 2) \]

### Step 6: Final Result
Thus, the value of the integral is:

\[ I = \ln(\sqrt{3} + 2) \]