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Câu hỏi: Cho hàm số $f\left( x \right)$ xác định và có đạo hàm ${f}'\left( x \right)$ liên tục trên đoạn $\left[ 1;3 \right]$ và $f\left( x \right)\ne 0$ với mọi $x\in \left[ 1;3 \right]$, đồng thời ${f}'\left( x \right){{\left( 1+f\left( x \right) \right)}^{2}}={{\left[ {{\left( f\left( x \right) \right)}^{2}}\left( x-1 \right) \right]}^{2}}$ và $f\left( 1 \right)=-1$. Khi đó $\int\limits_{1}^{3}{f\left( x \right)\text{d}x}$ là:
A. $-\ln 3$.
B. $\ln 3$.
C. $\ln 2$.
D. $-\ln 2$.$$
A. $-\ln 3$.
B. $\ln 3$.
C. $\ln 2$.
D. $-\ln 2$.$$
Ta có ${f}'\left( x \right){{\left( 1+f\left( x \right) \right)}^{2}}={{\left[ {{\left( f\left( x \right) \right)}^{2}}\left( x-1 \right) \right]}^{2}}$ $\Leftrightarrow \dfrac{{f}'\left( x \right){{\left( 1+f\left( x \right) \right)}^{2}}}{{{\left[ f\left( x \right) \right]}^{4}}}={{\left( x-1 \right)}^{2}}$.
Lấy nguyên hàm 2 vế ta được $\int{\dfrac{{f}'\left( x \right){{\left( 1+f\left( x \right) \right)}^{2}}}{{{\left[ f\left( x \right) \right]}^{4}}}\text{d}x}=\int{{{\left( x-1 \right)}^{2}}\text{d}x}$
$\Leftrightarrow \int{\dfrac{\left( 1+2f\left( x \right)+{{\left[ f\left( x \right) \right]}^{2}} \right){f}'\left( x \right)}{{{\left[ f\left( x \right) \right]}^{4}}}\text{d}x}=\int{{{\left( x-1 \right)}^{2}}\text{d}x}$
$\Leftrightarrow \int{\left( \dfrac{1}{{{\left[ f\left( x \right) \right]}^{4}}}+2\dfrac{1}{{{\left[ f\left( x \right) \right]}^{3}}}+\dfrac{1}{{{\left[ f\left( x \right) \right]}^{2}}} \right)\text{d}\left( f\left( x \right) \right)}=\dfrac{{{\left( x-1 \right)}^{3}}}{3}+C$
$\Leftrightarrow -\dfrac{1}{3{{\left[ f\left( x \right) \right]}^{3}}}-\dfrac{1}{{{\left[ f\left( x \right) \right]}^{2}}}-\dfrac{1}{f\left( x \right)}=\dfrac{{{\left( x-1 \right)}^{3}}}{3}+C$ $\Leftrightarrow -\dfrac{1+3f\left( x \right)+3{{\left[ f\left( x \right) \right]}^{2}}}{3{{\left[ f\left( x \right) \right]}^{3}}}=\dfrac{{{\left( x-1 \right)}^{3}}}{3}+C$
Mà $f\left( 1 \right)=-1$ nên $-\dfrac{1-3+3}{-3}=C\Rightarrow C=\dfrac{1}{3}$.
Suy ra $-\dfrac{1+3f\left( x \right)+3{{\left[ f\left( x \right) \right]}^{2}}}{3{{\left[ f\left( x \right) \right]}^{3}}}=\dfrac{{{\left( x-1 \right)}^{3}}}{3}+\dfrac{1}{3}$ $\Leftrightarrow $ $\dfrac{1+3f\left( x \right)+3{{\left[ f\left( x \right) \right]}^{2}}}{3{{\left[ f\left( x \right) \right]}^{3}}}+\dfrac{1}{3}=-\dfrac{{{\left( x-1 \right)}^{3}}}{3}$
$\Leftrightarrow \dfrac{{{\left( 1+f\left( x \right) \right)}^{3}}}{{{\left[ f\left( x \right) \right]}^{3}}}=-{{\left( x-1 \right)}^{3}}$ $\Leftrightarrow {{\left( 1+\dfrac{1}{f\left( x \right)} \right)}^{3}}={{\left( 1-x \right)}^{3}}$ $\Leftrightarrow f\left( x \right)=\dfrac{-1}{x}$.
Vậy $\int\limits_{1}^{3}{f\left( x \right)\text{d}x}=\left. \int\limits_{1}^{3}{\dfrac{-1}{x}\text{d}x=-\ln \left| x \right|} \right|_{1}^{3}=-\ln 3$.
Lấy nguyên hàm 2 vế ta được $\int{\dfrac{{f}'\left( x \right){{\left( 1+f\left( x \right) \right)}^{2}}}{{{\left[ f\left( x \right) \right]}^{4}}}\text{d}x}=\int{{{\left( x-1 \right)}^{2}}\text{d}x}$
$\Leftrightarrow \int{\dfrac{\left( 1+2f\left( x \right)+{{\left[ f\left( x \right) \right]}^{2}} \right){f}'\left( x \right)}{{{\left[ f\left( x \right) \right]}^{4}}}\text{d}x}=\int{{{\left( x-1 \right)}^{2}}\text{d}x}$
$\Leftrightarrow \int{\left( \dfrac{1}{{{\left[ f\left( x \right) \right]}^{4}}}+2\dfrac{1}{{{\left[ f\left( x \right) \right]}^{3}}}+\dfrac{1}{{{\left[ f\left( x \right) \right]}^{2}}} \right)\text{d}\left( f\left( x \right) \right)}=\dfrac{{{\left( x-1 \right)}^{3}}}{3}+C$
$\Leftrightarrow -\dfrac{1}{3{{\left[ f\left( x \right) \right]}^{3}}}-\dfrac{1}{{{\left[ f\left( x \right) \right]}^{2}}}-\dfrac{1}{f\left( x \right)}=\dfrac{{{\left( x-1 \right)}^{3}}}{3}+C$ $\Leftrightarrow -\dfrac{1+3f\left( x \right)+3{{\left[ f\left( x \right) \right]}^{2}}}{3{{\left[ f\left( x \right) \right]}^{3}}}=\dfrac{{{\left( x-1 \right)}^{3}}}{3}+C$
Mà $f\left( 1 \right)=-1$ nên $-\dfrac{1-3+3}{-3}=C\Rightarrow C=\dfrac{1}{3}$.
Suy ra $-\dfrac{1+3f\left( x \right)+3{{\left[ f\left( x \right) \right]}^{2}}}{3{{\left[ f\left( x \right) \right]}^{3}}}=\dfrac{{{\left( x-1 \right)}^{3}}}{3}+\dfrac{1}{3}$ $\Leftrightarrow $ $\dfrac{1+3f\left( x \right)+3{{\left[ f\left( x \right) \right]}^{2}}}{3{{\left[ f\left( x \right) \right]}^{3}}}+\dfrac{1}{3}=-\dfrac{{{\left( x-1 \right)}^{3}}}{3}$
$\Leftrightarrow \dfrac{{{\left( 1+f\left( x \right) \right)}^{3}}}{{{\left[ f\left( x \right) \right]}^{3}}}=-{{\left( x-1 \right)}^{3}}$ $\Leftrightarrow {{\left( 1+\dfrac{1}{f\left( x \right)} \right)}^{3}}={{\left( 1-x \right)}^{3}}$ $\Leftrightarrow f\left( x \right)=\dfrac{-1}{x}$.
Vậy $\int\limits_{1}^{3}{f\left( x \right)\text{d}x}=\left. \int\limits_{1}^{3}{\dfrac{-1}{x}\text{d}x=-\ln \left| x \right|} \right|_{1}^{3}=-\ln 3$.
Đáp án A.