Câu hỏi 12

\(\begin{array}{l}f'\left( x \right) = \dfrac{x}{{{x^2} + 1}}\\ \Rightarrow f\left( x \right) = \int\limits_{}^{} {f'\left( x \right)dx}  = \int\limits_{}^{} {\dfrac{{xdx}}{{{x^2} + 1}}}  = \dfrac{1}{2}\int\limits_{}^{} {\dfrac{{d\left( {{x^2} + 1} \right)}}{{{x^2} + 1}}}  = \dfrac{1}{2}\ln \left( {{x^2} + 1} \right) + C\\f\left( 0 \right) = 0 \Leftrightarrow \dfrac{1}{2}\ln 1 + C = 0 \Leftrightarrow C = 0 \Rightarrow f\left( x \right) = \dfrac{1}{2}\ln \left( {{x^2} + 1} \right)\\ \Rightarrow g\left( x \right) = 4xf\left( x \right) = 2x\ln \left( {{x^2} + 1} \right) \Rightarrow \int\limits_{}^{} {g\left( x \right)dx}  = \int\limits_{}^{} {2x\ln \left( {{x^2} + 1} \right)dx} \end{array}\)

Đặt \(t = {x^2} + 1 \Rightarrow dt = 2xdx\)

\(\begin{array}{l} \Rightarrow \int\limits_{}^{} {g\left( x \right)dx}  = \int\limits_{}^{} {\ln tdt}  = t\ln t - \int\limits_{}^{} {t.\dfrac{1}{t}dt}  = t\ln t - \int\limits_{}^{} {dt}  = t\ln t - t + C\\ = \left( {{x^2} + 1} \right)\ln \left( {{x^2} + 1} \right) - \left( {{x^2} + 1} \right) + C\end{array}\)

Đặt \( - 1 + C = c \Rightarrow \int\limits_{}^{} {g\left( x \right)dx}  = \left( {{x^2} + 1} \right)\ln \left( {{x^2} + 1} \right) - {x^2} + c\).

Chọn C