Google
Câu hỏi: Cho $F\left( x \right)=-\dfrac{1}{3{{x}^{3}}}$ là một nguyên hàm của hàm số $\dfrac{f\left( x \right)}{x}$. Nguyên hàm của hàm số ${f}'\left( x \right)\ln x$ bằng
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: ${F}'\left( x \right)=\dfrac{f\left( x \right)}{x}\Rightarrow f\left( x \right)=x.{F}'\left( x \right)=x.{{\left( -\dfrac{1}{3}.{{x}^{-3}} \right)}^{\prime }}=\dfrac{1}{{{x}^{3}}}={{x}^{-3}}$
$\Rightarrow {f}'\left( x \right)=-3{{x}^{-4}}\Rightarrow {f}'\left( x \right)\ln x=-3{{x}^{-4}}\ln x$. Do đó $\int{{f}'\left( x \right)\ln xdx}=\int{\left( -3{{x}^{-4}}\ln x \right)dx}$
Đặt $u=\ln x;dv={{x}^{-4}}dx\Rightarrow du=\dfrac{dx}{x};v=\dfrac{{{x}^{-3}}}{-3}.$
Nên $\int{{f}'\left( x \right)\ln xdx}=-3\int{\ln x.{{x}^{-4}}dx}=-3\left( \dfrac{\ln x}{-3{{x}^{3}}}+\int{\dfrac{{{x}^{-4}}}{3}dx} \right)=\dfrac{\ln x}{{{x}^{3}}}-\int{{{x}^{-4}}dx}=\dfrac{\ln x}{{{x}^{3}}}+\dfrac{1}{3{{x}^{3}}}+C.$
$\Rightarrow {f}'\left( x \right)=-3{{x}^{-4}}\Rightarrow {f}'\left( x \right)\ln x=-3{{x}^{-4}}\ln x$. Do đó $\int{{f}'\left( x \right)\ln xdx}=\int{\left( -3{{x}^{-4}}\ln x \right)dx}$
Đặt $u=\ln x;dv={{x}^{-4}}dx\Rightarrow du=\dfrac{dx}{x};v=\dfrac{{{x}^{-3}}}{-3}.$
Nên $\int{{f}'\left( x \right)\ln xdx}=-3\int{\ln x.{{x}^{-4}}dx}=-3\left( \dfrac{\ln x}{-3{{x}^{3}}}+\int{\dfrac{{{x}^{-4}}}{3}dx} \right)=\dfrac{\ln x}{{{x}^{3}}}-\int{{{x}^{-4}}dx}=\dfrac{\ln x}{{{x}^{3}}}+\dfrac{1}{3{{x}^{3}}}+C.$
Đáp án C.