Đáp án đúng: A
Giải chi tiết:\(\begin{array}{l}f\left( x \right) = {\left( {\dfrac{{{x^2} + 1}}{x}} \right)^2} = {\left( {x + \dfrac{1}{x}} \right)^2}\\\,\,\,\,\,\,\,\,\,\,\,\, = {x^2} + 2x.\dfrac{1}{x} + \dfrac{1}{{{x^2}}} = {x^2} + \dfrac{1}{{{x^2}}} + 2\\ \Rightarrow \int {f\left( x \right)dx} = \int {\left( {{x^2} + \dfrac{1}{{{x^2}}} + 2} \right)dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{{x^3}}}{3} - \dfrac{1}{x} + 2x + C\end{array}\)
Chọn A.