Đáp án:

\(x^2 (f(x))^2 + (2x - 1)f(x) = x f'(x) - 1\)

\(\Leftrightarrow x^2 (f(x))^2 + 2x f(x) + 1 = x f'(x) + f(x)\)

\(\Leftrightarrow (x f(x) + 1)^2 = (x f(x))'\)

\(\Leftrightarrow \frac{ (x f(x))'}{x f(x) + 1)^2} = 1 \Leftrightarrow \frac{(x f(x) + 1)'}{(x f(x) + 1)^2} = 1\)

\(\Leftrightarrow -\frac{1}{x f(x) + 1} = x + C\)

\(f(1) = -2 \Rightarrow C = 0\)

\(\Rightarrow x f(x) + 1 = -\frac{1}{x} \Rightarrow f(x) = \frac{1}{x} \left(-1 - \frac{1}{x}\right) = -\frac{1}{x} - \frac{1}{x^2}\)

\(\int_{1}^{2} f(x) dx = \int_{1}^{2} \left(-\frac{1}{x} - \frac{1}{x^2}\right) dx = -\ln{x} + \frac{1}{x} \Big|_{1}^{2}\)

\(= -\ln{2} + \frac{1}{2} - 1 = -\frac{1}{2} - \ln{2}\Rightarrow \boxed{B}\)

Gợi ý: \( (x f(x))' = f(x) + x f'(x)\)