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Câu hỏi: Biết rằng ${F\left( x \right)=-\dfrac{1}{{{x}^{5}}}}$ là một nguyên hàm của hàm số ${\dfrac{f\left( x \right)}{x}}$. Tmf nguyên hàm của hàm số ${{f}'\left( x \right)\ln x}$.
A. ${\int{{f}'\left( x \right)dx}=\dfrac{\ln x}{25{{x}^{5}}}+\dfrac{1}{5{{x}^{5}}}+C}$.
B. ${\int{{f}'\left( x \right)dx}=\dfrac{5\ln x}{{{x}^{5}}}+\dfrac{1}{{{x}^{5}}}+C}$.
C. ${\int{{f}'\left( x \right)dx}=\dfrac{\ln x}{5{{x}^{5}}}-\dfrac{1}{5{{x}^{5}}}+C}$.
D. ${\int{{f}'\left( x \right)dx}=\dfrac{5\ln x}{{{x}^{5}}}-\dfrac{1}{5{{x}^{5}}}+C}$.
A. ${\int{{f}'\left( x \right)dx}=\dfrac{\ln x}{25{{x}^{5}}}+\dfrac{1}{5{{x}^{5}}}+C}$.
B. ${\int{{f}'\left( x \right)dx}=\dfrac{5\ln x}{{{x}^{5}}}+\dfrac{1}{{{x}^{5}}}+C}$.
C. ${\int{{f}'\left( x \right)dx}=\dfrac{\ln x}{5{{x}^{5}}}-\dfrac{1}{5{{x}^{5}}}+C}$.
D. ${\int{{f}'\left( x \right)dx}=\dfrac{5\ln x}{{{x}^{5}}}-\dfrac{1}{5{{x}^{5}}}+C}$.
Ta có: $F(x)=-\dfrac{1}{{{x}^{5}}}\Rightarrow {{F}^{\prime }}(x)=\dfrac{5}{{{x}^{6}}}\Rightarrow f(x)=\dfrac{5}{{{x}^{5}}}\Rightarrow {{f}^{\prime }}(x)=-\dfrac{25}{{{x}^{6}}}$
Xét $\int{{{f}^{\prime }}}(x)\ln xdx$
Đặt$\left\{ \begin{array}{*{35}{l}}
u=\ln x \\
dv={{f}^{\prime }}(x)dx \\
\end{array}\Rightarrow \left\{ \begin{array}{*{35}{l}}
du=\dfrac{1}{x}dx \\
v=f(x) \\
\end{array} \right. \right. $. Do đó$ \int{{{f}^{\prime }}}(x)\ln xdx=f(x)\ln x-\int{f}(x)\cdot \dfrac{1}{x}dx$
$=\dfrac{5\ln x}{{{x}^{5}}}-\int{\dfrac{5}{{{x}^{6}}}}dx=\dfrac{5\ln x}{{{x}^{5}}}-5\int{{{x}^{-6}}}dx=\dfrac{5\ln x}{{{x}^{5}}}+\dfrac{1}{{{x}^{5}}}+C$
Xét $\int{{{f}^{\prime }}}(x)\ln xdx$
Đặt$\left\{ \begin{array}{*{35}{l}}
u=\ln x \\
dv={{f}^{\prime }}(x)dx \\
\end{array}\Rightarrow \left\{ \begin{array}{*{35}{l}}
du=\dfrac{1}{x}dx \\
v=f(x) \\
\end{array} \right. \right. $. Do đó$ \int{{{f}^{\prime }}}(x)\ln xdx=f(x)\ln x-\int{f}(x)\cdot \dfrac{1}{x}dx$
$=\dfrac{5\ln x}{{{x}^{5}}}-\int{\dfrac{5}{{{x}^{6}}}}dx=\dfrac{5\ln x}{{{x}^{5}}}-5\int{{{x}^{-6}}}dx=\dfrac{5\ln x}{{{x}^{5}}}+\dfrac{1}{{{x}^{5}}}+C$
Đáp án B.