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Câu hỏi: Cho $F\left( x \right)=3{{x}^{2}}$ là một nguyên hàm của hàm số $\dfrac{f\left( x \right)}{x}.$ Tích phân $\int\limits_{1}^{2}{{f}'\left( x \right)\ln xdx}$ bằng
A. $-3+12\ln 2.$
B. $-9+24\ln 2.$
C. $3+6\ln 2.$
D. $1+9\ln 2.$
A. $-3+12\ln 2.$
B. $-9+24\ln 2.$
C. $3+6\ln 2.$
D. $1+9\ln 2.$
Ta có $\dfrac{f\left( x \right)}{x}={F}'\left( x \right)=6x\Rightarrow f\left( x \right)=6{{x}^{2}}\Rightarrow {f}'\left( x \right)=12x$
$\Rightarrow \int\limits_{1}^{2}{{f}'\left( x \right)\ln xdx}=\int\limits_{1}^{2}{12x.\ln xdx}=6\int\limits_{1}^{2}{\ln xd\left( {{x}^{2}} \right)}=\left. 6{{x}^{2}}\ln x \right|_{1}^{2}-6\int\limits_{1}^{2}{{{x}^{2}}d\left( \ln x \right)}$
$=24\ln 2-6\int\limits_{1}^{2}{{{x}^{2}}.\dfrac{1}{x}dx}=24\ln 2-\left. 3{{x}^{2}} \right|_{0}^{1}=-9+24\ln 2$.
$\Rightarrow \int\limits_{1}^{2}{{f}'\left( x \right)\ln xdx}=\int\limits_{1}^{2}{12x.\ln xdx}=6\int\limits_{1}^{2}{\ln xd\left( {{x}^{2}} \right)}=\left. 6{{x}^{2}}\ln x \right|_{1}^{2}-6\int\limits_{1}^{2}{{{x}^{2}}d\left( \ln x \right)}$
$=24\ln 2-6\int\limits_{1}^{2}{{{x}^{2}}.\dfrac{1}{x}dx}=24\ln 2-\left. 3{{x}^{2}} \right|_{0}^{1}=-9+24\ln 2$.
Đáp án B.