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Câu hỏi: Cho hàm số $y=f\left( x \right)$ xác định và liên tục trên $\mathbb{R}\backslash \left\{ 0 \right\}$ thỏa mãn ${{\left[ x.f\left( x \right) \right]}^{2}}+\left( 2x-1 \right).f\left( x \right)=x.{f}'\left( x \right)-1$ và $f\left( 2 \right)=-\dfrac{3}{4}$. Tích phân $\int\limits_{1}^{9}{f\left( x \right)dx}$ bằng
A. $\dfrac{8}{9}-2\ln 3$
B. $-\dfrac{8}{9}-2\ln 3$
C. $\dfrac{2}{9}-\ln 3$
D. $-\dfrac{2}{9}-\ln 3$
A. $\dfrac{8}{9}-2\ln 3$
B. $-\dfrac{8}{9}-2\ln 3$
C. $\dfrac{2}{9}-\ln 3$
D. $-\dfrac{2}{9}-\ln 3$
Ta có ${{\left( x.f\left( x \right)+1 \right)}^{2}}=f\left( x \right)+x.{f}'\left( x \right)\Rightarrow {{\left( x.f\left( x \right)+1 \right)}^{2}}={{\left[ x.f\left( x \right)+1 \right]}^{\prime }}$
Đặt $g\left( x \right)=x.f\left( x \right)+1\Rightarrow {{\left[ g\left( x \right) \right]}^{2}}={g}'\left( x \right)\Rightarrow \int{\dfrac{{g}'\left( x \right)}{{{\left[ g\left( x \right) \right]}^{2}}}dx}=x+{{C}_{1}}$
$\Rightarrow \int{\dfrac{1}{{{\left[ g\left( x \right) \right]}^{2}}}d\left[ g\left( x \right) \right]}=x+{{C}_{1}}\Rightarrow -\dfrac{1}{g\left( x \right)}=x+{{C}_{2}}\Rightarrow -\dfrac{1}{x.f\left( x \right)+1}=x+{{C}_{2}}$
$\Rightarrow x.f\left( x \right)=-\dfrac{1}{x}-1\Rightarrow f\left( x \right)=-\dfrac{1}{{{x}^{2}}}-\dfrac{1}{x}$
$\Rightarrow \int\limits_{1}^{9}{f\left( x \right)dx}=\int\limits_{1}^{9}{\left( -\dfrac{1}{{{x}^{2}}}-\dfrac{1}{x} \right)dx}=\left( \dfrac{1}{x}-\ln \left| x \right| \right)\left| \begin{aligned}
& ^{9} \\
& _{1} \\
\end{aligned} \right.=\left( \dfrac{1}{9}-\ln 9 \right)-1=-\dfrac{8}{9}-2$
Chọn B.
Đặt $g\left( x \right)=x.f\left( x \right)+1\Rightarrow {{\left[ g\left( x \right) \right]}^{2}}={g}'\left( x \right)\Rightarrow \int{\dfrac{{g}'\left( x \right)}{{{\left[ g\left( x \right) \right]}^{2}}}dx}=x+{{C}_{1}}$
$\Rightarrow \int{\dfrac{1}{{{\left[ g\left( x \right) \right]}^{2}}}d\left[ g\left( x \right) \right]}=x+{{C}_{1}}\Rightarrow -\dfrac{1}{g\left( x \right)}=x+{{C}_{2}}\Rightarrow -\dfrac{1}{x.f\left( x \right)+1}=x+{{C}_{2}}$
$\Rightarrow x.f\left( x \right)=-\dfrac{1}{x}-1\Rightarrow f\left( x \right)=-\dfrac{1}{{{x}^{2}}}-\dfrac{1}{x}$
$\Rightarrow \int\limits_{1}^{9}{f\left( x \right)dx}=\int\limits_{1}^{9}{\left( -\dfrac{1}{{{x}^{2}}}-\dfrac{1}{x} \right)dx}=\left( \dfrac{1}{x}-\ln \left| x \right| \right)\left| \begin{aligned}
& ^{9} \\
& _{1} \\
\end{aligned} \right.=\left( \dfrac{1}{9}-\ln 9 \right)-1=-\dfrac{8}{9}-2$
Chọn B.
Đáp án B.