1. **State the problem:** Evaluate the definite integral $$\int_0^{\frac{\pi}{2}} \frac{x^2}{(x-1)^2} \, dx.$$\n\n2. **Rewrite the integrand:** We have the function $$f(x) = \frac{x^2}{(x-1)^2}.$$ To simplify, perform polynomial division or rewrite numerator in terms of $(x-1)$:\n\n$$x^2 = (x-1+1)^2 = (x-1)^2 + 2(x-1) + 1.$$\n\nSo,\n$$\frac{x^2}{(x-1)^2} = \frac{(x-1)^2 + 2(x-1) + 1}{(x-1)^2} = 1 + \frac{2(x-1)}{(x-1)^2} + \frac{1}{(x-1)^2} = 1 + \frac{2}{x-1} + \frac{1}{(x-1)^2}.$$\n\n3. **Rewrite the integral:**\n$$\int_0^{\frac{\pi}{2}} \frac{x^2}{(x-1)^2} \, dx = \int_0^{\frac{\pi}{2}} \left(1 + \frac{2}{x-1} + \frac{1}{(x-1)^2}\right) dx.$$\n\n4. **Integrate term-by-term:**\n- $$\int 1 \, dx = x.$$\n- $$\int \frac{2}{x-1} \, dx = 2 \ln|x-1|.$$\n- $$\int \frac{1}{(x-1)^2} \, dx = \int (x-1)^{-2} \, dx = -\frac{1}{x-1}.$$\n\n5. **Combine the antiderivative:**\n$$F(x) = x + 2 \ln|x-1| - \frac{1}{x-1} + C.$$\n\n6. **Evaluate the definite integral:**\n$$\int_0^{\frac{\pi}{2}} \frac{x^2}{(x-1)^2} \, dx = F\left(\frac{\pi}{2}\right) - F(0) = \left(\frac{\pi}{2} + 2 \ln\left|\frac{\pi}{2} - 1\right| - \frac{1}{\frac{\pi}{2} - 1}\right) - \left(0 + 2 \ln|0 - 1| - \frac{1}{0 - 1}\right).$$\n\n7. **Simplify the evaluation:**\n- Note that $\ln|0-1| = \ln 1 = 0.$\n- Also, $\frac{1}{0-1} = \frac{1}{-1} = -1.$\n\nSo,\n$$F\left(\frac{\pi}{2}\right) - F(0) = \frac{\pi}{2} + 2 \ln\left(\frac{\pi}{2} - 1\right) - \frac{1}{\frac{\pi}{2} - 1} - (0 + 0 - (-1)) = \frac{\pi}{2} + 2 \ln\left(\frac{\pi}{2} - 1\right) - \frac{1}{\frac{\pi}{2} - 1} + 1.$$\n\n**Final answer:**\n$$\boxed{\int_0^{\frac{\pi}{2}} \frac{x^2}{(x-1)^2} \, dx = \frac{\pi}{2} + 1 + 2 \ln\left(\frac{\pi}{2} - 1\right) - \frac{1}{\frac{\pi}{2} - 1}}.$$