Google
Câu hỏi: Cho hàm số $f\left( x \right)$ xác định và có đạo hàm trên $\left( 0 ;+\infty \right)$ thỏa mãn $f\left( 1 \right)=\dfrac{4}{e}$ và
$\left( x+1 \right)f\left( x \right)+x{f}'\left( x \right)=\left( 2x+1 \right){{e}^{-x}}$ với mọi $x>0$. Tính $\int\limits_{1}^{2}{{{e}^{x}}f\left( x \right)dx}$.
A. $4-\ln 4.$
B. $\dfrac{5}{2}-2\ln 2.$
C. $4+\ln 4.$
D. $\dfrac{5}{2}+2\ln 2.$
$\left( x+1 \right)f\left( x \right)+x{f}'\left( x \right)=\left( 2x+1 \right){{e}^{-x}}$ với mọi $x>0$. Tính $\int\limits_{1}^{2}{{{e}^{x}}f\left( x \right)dx}$.
A. $4-\ln 4.$
B. $\dfrac{5}{2}-2\ln 2.$
C. $4+\ln 4.$
D. $\dfrac{5}{2}+2\ln 2.$
Ta có $\left( x+1 \right)f\left( x \right)+x{f}'\left( x \right)=\left( 2x+1 \right){{e}^{-x}}$ $\Leftrightarrow \left( x+1 \right){{e}^{x}}f\left( x \right)+{{e}^{x}}x{f}'\left( x \right)=2x+1$
$\Leftrightarrow {{\left[ x{{e}^{x}}f\left( x \right) \right]}^{\prime }}=2x+1$ $\Rightarrow \int{{{\left[ x{{e}^{x}}f\left( x \right) \right]}^{\prime }}}dx=\int{\left( 2x+1 \right)}dx$ $\Leftrightarrow x{{e}^{x}}f\left( x \right)={{x}^{2}}+x+C$.
Vì $f\left( 1 \right)=\dfrac{4}{e}$ nên $C=ef\left( 1 \right)-2=2$. Suy ra ${{e}^{x}}f\left( x \right)=x+1+\dfrac{2}{x}$.
Khi đó $\int\limits_{1}^{2}{{{e}^{x}}f\left( x \right)dx}=\int\limits_{1}^{2}{\left( x+1+\dfrac{2}{x} \right)}dx=\left( \dfrac{{{x}^{2}}}{2}+x+2\ln \left| x \right| \right)\left| _{1}^{2} \right.=$ $\dfrac{5}{2}+2\ln 2.$
$\Leftrightarrow {{\left[ x{{e}^{x}}f\left( x \right) \right]}^{\prime }}=2x+1$ $\Rightarrow \int{{{\left[ x{{e}^{x}}f\left( x \right) \right]}^{\prime }}}dx=\int{\left( 2x+1 \right)}dx$ $\Leftrightarrow x{{e}^{x}}f\left( x \right)={{x}^{2}}+x+C$.
Vì $f\left( 1 \right)=\dfrac{4}{e}$ nên $C=ef\left( 1 \right)-2=2$. Suy ra ${{e}^{x}}f\left( x \right)=x+1+\dfrac{2}{x}$.
Khi đó $\int\limits_{1}^{2}{{{e}^{x}}f\left( x \right)dx}=\int\limits_{1}^{2}{\left( x+1+\dfrac{2}{x} \right)}dx=\left( \dfrac{{{x}^{2}}}{2}+x+2\ln \left| x \right| \right)\left| _{1}^{2} \right.=$ $\dfrac{5}{2}+2\ln 2.$
Đáp án D.
