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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên $\mathbb{R}$ và thỏa mãn các điều kiện $f\left( 0 \right)=-2$ và $\left( {{x}^{2}}+1 \right){f}'\left( x \right)+xf\left( x \right)=-x$, $\forall x\in \mathbb{R}$. Tính tích phân $I=\int\limits_{0}^{\sqrt{3}}{xf\left( x \right)\text{d}x}$.
A. $I=-\dfrac{4}{3}$.
B. $I=-\dfrac{1}{2}$.
C. $I=-\dfrac{3}{2}$.
D. $I=-\dfrac{5}{2}$.
A. $I=-\dfrac{4}{3}$.
B. $I=-\dfrac{1}{2}$.
C. $I=-\dfrac{3}{2}$.
D. $I=-\dfrac{5}{2}$.
Ta có: $\left( {{x}^{2}}+1 \right){f}'\left( x \right)+xf\left( x \right)=-x$ $\Leftrightarrow \sqrt{{{x}^{2}}+1}.{f}'\left( x \right)+\dfrac{x}{\sqrt{{{x}^{2}}+1}}.f\left( x \right)=-\dfrac{x}{\sqrt{{{x}^{2}}+1}}$
$\Leftrightarrow {{\left[ \sqrt{{{x}^{2}}+1}.f\left( x \right) \right]}^{\prime }}={{\left( -\sqrt{{{x}^{2}}+1} \right)}^{\prime }}$ $\Leftrightarrow \sqrt{{{x}^{2}}+1}.f\left( x \right)=-\sqrt{{{x}^{2}}+1}+C$ $\Leftrightarrow f\left( x \right)=-1+\dfrac{C}{\sqrt{{{x}^{2}}+1}}$.
Vì $f\left( 0 \right)=-2$ nên $-2=-1+\dfrac{C}{\sqrt{{{0}^{2}}+1}}\Rightarrow C=-1$. Do đó $f\left( x \right)=-1-\dfrac{1}{\sqrt{{{x}^{2}}+1}}$.
Khi đó
$I=\int\limits_{0}^{\sqrt{3}}{xf\left( x \right)\text{d}x}=\int\limits_{0}^{\sqrt{3}}{\left[ -x-\dfrac{x}{\sqrt{{{x}^{2}}+1}} \right]\text{d}x}=\left. \left[ -\dfrac{1}{2}{{x}^{2}}-\sqrt{{{x}^{2}}+1} \right] \right|_{0}^{\sqrt{3}}=\left( -\dfrac{3}{2}-\sqrt{3+1} \right)-\left( -0-1 \right)=-\dfrac{5}{2}$.
$\Leftrightarrow {{\left[ \sqrt{{{x}^{2}}+1}.f\left( x \right) \right]}^{\prime }}={{\left( -\sqrt{{{x}^{2}}+1} \right)}^{\prime }}$ $\Leftrightarrow \sqrt{{{x}^{2}}+1}.f\left( x \right)=-\sqrt{{{x}^{2}}+1}+C$ $\Leftrightarrow f\left( x \right)=-1+\dfrac{C}{\sqrt{{{x}^{2}}+1}}$.
Vì $f\left( 0 \right)=-2$ nên $-2=-1+\dfrac{C}{\sqrt{{{0}^{2}}+1}}\Rightarrow C=-1$. Do đó $f\left( x \right)=-1-\dfrac{1}{\sqrt{{{x}^{2}}+1}}$.
Khi đó
$I=\int\limits_{0}^{\sqrt{3}}{xf\left( x \right)\text{d}x}=\int\limits_{0}^{\sqrt{3}}{\left[ -x-\dfrac{x}{\sqrt{{{x}^{2}}+1}} \right]\text{d}x}=\left. \left[ -\dfrac{1}{2}{{x}^{2}}-\sqrt{{{x}^{2}}+1} \right] \right|_{0}^{\sqrt{3}}=\left( -\dfrac{3}{2}-\sqrt{3+1} \right)-\left( -0-1 \right)=-\dfrac{5}{2}$.
Đáp án D.