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Câu hỏi: Cho $F\left( x \right)=-\dfrac{1}{3{{x}^{3}}}$ là một nguyên hàm của hàm số $\dfrac{f\left( x \right)}{x}.$ Tìm nguyên hàm của hàm số ${f}'\left( x \right)\ln x.$
A. $\int{{f}'\left( x \right)\ln xdx}=\dfrac{\ln x}{{{x}^{3}}}+\dfrac{1}{5{{x}^{5}}}+C.$
B. $\int{{f}'\left( x \right)\ln xdx}=-\dfrac{\ln x}{{{x}^{3}}}+\dfrac{1}{3{{x}^{3}}}+C.$
C. $\int{{f}'\left( x \right)\ln xdx}=\dfrac{\ln x}{{{x}^{3}}}+\dfrac{1}{3{{x}^{3}}}+C.$
D. $\int{{f}'\left( x \right)\ln xdx}=\dfrac{\ln x}{{{x}^{3}}}-\dfrac{1}{5{{x}^{5}}}+C.$
A. $\int{{f}'\left( x \right)\ln xdx}=\dfrac{\ln x}{{{x}^{3}}}+\dfrac{1}{5{{x}^{5}}}+C.$
B. $\int{{f}'\left( x \right)\ln xdx}=-\dfrac{\ln x}{{{x}^{3}}}+\dfrac{1}{3{{x}^{3}}}+C.$
C. $\int{{f}'\left( x \right)\ln xdx}=\dfrac{\ln x}{{{x}^{3}}}+\dfrac{1}{3{{x}^{3}}}+C.$
D. $\int{{f}'\left( x \right)\ln xdx}=\dfrac{\ln x}{{{x}^{3}}}-\dfrac{1}{5{{x}^{5}}}+C.$
Ta có $\dfrac{f\left( x \right)}{x}={F}'\left( x \right)={{\left( \dfrac{-1}{3{{x}^{3}}} \right)}^{\prime }}=\dfrac{1}{{{x}^{4}}}\Rightarrow f\left( x \right)=\dfrac{1}{{{x}^{3}}}.$
Do đó $\int{{f}'\left( x \right)\ln xdx}=\int{\dfrac{-3}{{{x}^{4}}}\ln xdx}.$
Đặt $\left\{ \begin{aligned}
& u=\ln x \\
& dv=\dfrac{-3}{{{x}^{4}}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{x}dx \\
& v=\dfrac{1}{{{x}^{3}}} \\
\end{aligned} \right.. $ Suy ra $ \int{\dfrac{-3}{{{x}^{4}}}\ln xdx}=\dfrac{\ln x}{{{x}^{3}}}-\int{\dfrac{1}{{{x}^{4}}}dx}=\dfrac{\ln x}{{{x}^{3}}}+\dfrac{1}{3{{x}^{3}}}+C.$
Do đó $\int{{f}'\left( x \right)\ln xdx}=\int{\dfrac{-3}{{{x}^{4}}}\ln xdx}.$
Đặt $\left\{ \begin{aligned}
& u=\ln x \\
& dv=\dfrac{-3}{{{x}^{4}}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{x}dx \\
& v=\dfrac{1}{{{x}^{3}}} \\
\end{aligned} \right.. $ Suy ra $ \int{\dfrac{-3}{{{x}^{4}}}\ln xdx}=\dfrac{\ln x}{{{x}^{3}}}-\int{\dfrac{1}{{{x}^{4}}}dx}=\dfrac{\ln x}{{{x}^{3}}}+\dfrac{1}{3{{x}^{3}}}+C.$
Đáp án C.