1. The problem is to evaluate the integral $$\int_2^5 \frac{x}{(x-1)^{1|2|}} \, dx$$. 2. First, interpret the exponent: $1|2|$ means $1 \times |2| = 1 \times 2 = 2$. 3. So the integral becomes $$\int_2^5 \frac{x}{(x-1)^2} \, dx$$. 4. To solve this integral, use substitution or rewrite the integrand: $$\frac{x}{(x-1)^2} = \frac{(x-1)+1}{(x-1)^2} = \frac{x-1}{(x-1)^2} + \frac{1}{(x-1)^2} = \frac{1}{x-1} + (x-1)^{-2}$$ 5. Now the integral is: $$\int_2^5 \left( \frac{1}{x-1} + (x-1)^{-2} \right) dx = \int_2^5 \frac{1}{x-1} dx + \int_2^5 (x-1)^{-2} dx$$ 6. Evaluate each integral separately: - $$\int \frac{1}{x-1} dx = \ln|x-1| + C$$ - $$\int (x-1)^{-2} dx = \int u^{-2} du = -u^{-1} + C = -\frac{1}{x-1} + C$$ 7. So the definite integral is: $$\left[ \ln|x-1| - \frac{1}{x-1} \right]_2^5 = \left( \ln|5-1| - \frac{1}{5-1} \right) - \left( \ln|2-1| - \frac{1}{2-1} \right)$$ 8. Calculate the values: - $\ln 4 - \frac{1}{4} - (\ln 1 - 1) = \ln 4 - \frac{1}{4} - 0 + 1 = \ln 4 + 1 - \frac{1}{4}$ 9. Final answer: $$\boxed{\ln 4 + \frac{3}{4}}$$