To solve the given integral:

$\int_2^3 \frac{dx}{(x-1)^2 \sqrt{x^2 - 2x}}$

Steps to Solve:

1. Simplify the Radicand: Rewrite $x^2 - 2x$ as $x(x - 2)$, so the integrand becomes: $\int_2^3 \frac{dx}{(x-1)^2 \sqrt{x(x - 2)}}.$

2. Substitution: Let $u = x - 1$, so $x = u + 1$ and $dx = du$. The limits of integration change as follows:

   • When $x = 2$, $u = 1$,
   • When $x = 3$, $u = 2$.

   Substitute into the integral: $\int_2^3 \frac{dx}{(x-1)^2 \sqrt{x(x - 2)}} = \int_1^2 \frac{du}{u^2 \sqrt{(u + 1)(u - 1)}}.$

3. Simplify Further: The square root simplifies as: $\sqrt{(u + 1)(u - 1)} = \sqrt{u^2 - 1}.$

   The integral now becomes: $\int_1^2 \frac{du}{u^2 \sqrt{u^2 - 1}}.$

4. Use Trigonometric Substitution: Let $u = \sec\theta$, so $du = \sec\theta\tan\theta \, d\theta$ and $\sqrt{u^2 - 1} = \tan\theta$. The limits of integration change:

   • When $u = 1$, $\theta = 0$,
   • When $u = 2$, $\theta = \sec^{-1}(2)$.

   Substituting these, the integral becomes: $\int_0^{\sec^{-1}(2)} \frac{\sec\theta\tan\theta \, d\theta}{\sec^2\theta \tan\theta} = \int_0^{\sec^{-1}(2)} \frac{d\theta}{\sec\theta}.$

5. Simplify: Since $\frac{1}{\sec\theta} = \cos\theta$, the integral simplifies to: $\int_0^{\sec^{-1}(2)} \cos\theta \, d\theta.$

6. Evaluate the Integral: The integral of $\cos\theta$ is $\sin\theta$. Thus: $\int_0^{\sec^{-1}(2)} \cos\theta \, d\theta = \sin\theta \Big|_0^{\sec^{-1}(2)}.$

   Substitute the limits:

   • At $\theta = \sec^{-1}(2)$, $\sin\theta = \frac{\sqrt{3}}{2}$,
   • At $\theta = 0$, $\sin\theta = 0$.

   So the result is: $\frac{\sqrt{3}}{2} - 0 = \frac{\sqrt{3}}{2}.$

Final Answer:

$\boxed{\frac{\sqrt{3}}{2}}$

Would you like further clarification or more detailed steps? Here are related questions to expand your understanding:

1. What is the importance of substitution in evaluating complex integrals?
2. How does trigonometric substitution simplify square roots in integrals?
3. Could other substitutions (e.g., hyperbolic) solve this problem differently?
4. How can the limits of integration be verified after substitution?
5. What are the conditions under which trigonometric substitution is ideal?

Tip: Always verify that your substitution simplifies both the integr