Google
Câu hỏi: Cho hàm số $f\left( x \right)$ nhận giá trị dương và có đạo hàm liên tục trên $\left[ 1;\ e \right]$. Biết $f\left( 1 \right)=1$ và $x.f\left( x \right).{f}'\left( x \right)={{x}^{2}}+{{f}^{2}}\left( x \right)$ với mọi $x\in \left[ 1;\ e \right].$ Khi đó, $\int\limits_{1}^{e}{\dfrac{f\left( x \right)}{{{x}^{2}}}dx}$ bằng
A. $\dfrac{2}{3}$.
B. $\dfrac{3\sqrt{3}-1}{3}$.
C. $\dfrac{\sqrt{3}-1}{3}$.
D. $\sqrt{3}$.
A. $\dfrac{2}{3}$.
B. $\dfrac{3\sqrt{3}-1}{3}$.
C. $\dfrac{\sqrt{3}-1}{3}$.
D. $\sqrt{3}$.
Với $x\in \left[ 1;e \right]$ :
$x.f\left( x \right).{f}'\left( x \right)={{x}^{2}}+{{f}^{2}}\left( x \right)\Leftrightarrow 2{{x}^{2}}.f\left( x \right).{f}'\left( x \right)-2x.{{f}^{2}}\left( x \right)=2{{x}^{3}}$
$\Leftrightarrow \dfrac{{{x}^{2}}.{{\left( {{f}^{2}}\left( x \right) \right)}^{\prime }}-{{\left( {{x}^{2}} \right)}^{\prime }}{{f}^{2}}\left( x \right)}{{{x}^{4}}}=\dfrac{2}{x}\Leftrightarrow {{\left( \dfrac{{{f}^{2}}\left( x \right)}{{{x}^{2}}} \right)}^{\prime }}=\dfrac{2}{x}$
$\Leftrightarrow \int{{{\left( \dfrac{{{f}^{2}}\left( x \right)}{{{x}^{2}}} \right)}^{\prime }}dx}=\int{\dfrac{2}{x}dx}\Leftrightarrow \dfrac{{{f}^{2}}\left( x \right)}{{{x}^{2}}}=2\ln \left| x \right|+C$.
Thay $x=1$ và chú ý $f\left( 1 \right)=1$ ta được $\dfrac{{{f}^{2}}\left( 1 \right)}{{{1}^{2}}}=2\ln \left| 1 \right|+C\Leftrightarrow C=1.$
Do đó, trên $\left[ 1;\ e \right]$ : $\dfrac{{{f}^{2}}\left( x \right)}{{{x}^{2}}}=2\ln \left| x \right|+1\Leftrightarrow {{f}^{2}}\left( x \right)={{x}^{2}}\left( 2\ln x+1 \right)\Leftrightarrow f\left( x \right)=x\sqrt{2\ln x+1}.$
Và $I=\int\limits_{1}^{e}{\dfrac{f\left( x \right)}{{{x}^{2}}}dx}=\int\limits_{1}^{e}{\dfrac{\sqrt{2\ln x+1}}{x}}dx=\int\limits_{1}^{e}{\sqrt{2\ln x+1}d\left( \ln x \right)}=\int\limits_{0}^{1}{\sqrt{2t+1}dt}$
Đặt $u=\sqrt{2t+1}\Rightarrow {{u}^{2}}=2t+1\Rightarrow 2udu=2dt$ nên $I=\int\limits_{1}^{\sqrt{3}}{{{u}^{2}}du=\dfrac{3\sqrt{3}-1}{3}.}$
$x.f\left( x \right).{f}'\left( x \right)={{x}^{2}}+{{f}^{2}}\left( x \right)\Leftrightarrow 2{{x}^{2}}.f\left( x \right).{f}'\left( x \right)-2x.{{f}^{2}}\left( x \right)=2{{x}^{3}}$
$\Leftrightarrow \dfrac{{{x}^{2}}.{{\left( {{f}^{2}}\left( x \right) \right)}^{\prime }}-{{\left( {{x}^{2}} \right)}^{\prime }}{{f}^{2}}\left( x \right)}{{{x}^{4}}}=\dfrac{2}{x}\Leftrightarrow {{\left( \dfrac{{{f}^{2}}\left( x \right)}{{{x}^{2}}} \right)}^{\prime }}=\dfrac{2}{x}$
$\Leftrightarrow \int{{{\left( \dfrac{{{f}^{2}}\left( x \right)}{{{x}^{2}}} \right)}^{\prime }}dx}=\int{\dfrac{2}{x}dx}\Leftrightarrow \dfrac{{{f}^{2}}\left( x \right)}{{{x}^{2}}}=2\ln \left| x \right|+C$.
Thay $x=1$ và chú ý $f\left( 1 \right)=1$ ta được $\dfrac{{{f}^{2}}\left( 1 \right)}{{{1}^{2}}}=2\ln \left| 1 \right|+C\Leftrightarrow C=1.$
Do đó, trên $\left[ 1;\ e \right]$ : $\dfrac{{{f}^{2}}\left( x \right)}{{{x}^{2}}}=2\ln \left| x \right|+1\Leftrightarrow {{f}^{2}}\left( x \right)={{x}^{2}}\left( 2\ln x+1 \right)\Leftrightarrow f\left( x \right)=x\sqrt{2\ln x+1}.$
Và $I=\int\limits_{1}^{e}{\dfrac{f\left( x \right)}{{{x}^{2}}}dx}=\int\limits_{1}^{e}{\dfrac{\sqrt{2\ln x+1}}{x}}dx=\int\limits_{1}^{e}{\sqrt{2\ln x+1}d\left( \ln x \right)}=\int\limits_{0}^{1}{\sqrt{2t+1}dt}$
Đặt $u=\sqrt{2t+1}\Rightarrow {{u}^{2}}=2t+1\Rightarrow 2udu=2dt$ nên $I=\int\limits_{1}^{\sqrt{3}}{{{u}^{2}}du=\dfrac{3\sqrt{3}-1}{3}.}$
Đáp án B.