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Câu hỏi: Cho hàm số $f\left(x \right)$ có $f\left(1 \right)={{\text{e}}^{2}}$ và ${f}'\left(x \right)=\frac{2x-1}{{{x}^{2}}}{{\text{e}}^{2x}}$, $\forall x\ne 0$. Khi đó $\int\limits_{1}^{\ln 3}{xf\left(x \right)} \text{d}x$ bằng
A. $6-{{\text{e}}^{2}}$.
B. $\frac{6-{{\text{e}}^{2}}}{2}$.
C. $9-{{\text{e}}^{2}}$.
D. $\frac{9-{{\text{e}}^{2}}}{2}$.
A. $6-{{\text{e}}^{2}}$.
B. $\frac{6-{{\text{e}}^{2}}}{2}$.
C. $9-{{\text{e}}^{2}}$.
D. $\frac{9-{{\text{e}}^{2}}}{2}$.
+ ${f}'\left( x \right)=\frac{2x-1}{{{x}^{2}}}{{\text{e}}^{2x}}=\frac{2}{x}{{\text{e}}^{2x}}-\frac{1}{{{x}^{2}}}{{\text{e}}^{2x}}=\frac{1}{x}{{\left( {{\text{e}}^{2x}} \right)}^{\prime }}+{{\left( \frac{1}{x} \right)}^{\prime }}{{\text{e}}^{2x}}$
$\Rightarrow {f}'\left( x \right)={{\left( {{\text{e}}^{2x}}.\frac{1}{x} \right)}^{\prime }}\Rightarrow f\left( x \right)={{\text{e}}^{2x}}.\frac{1}{x}+C$.
+ Do $f\left( 1 \right)={{\text{e}}^{2}}$ nên ${{\text{e}}^{2.1}}.\frac{1}{1}+C={{\text{e}}^{2}}\Leftrightarrow C=0$.
+ Vậy $f\left( x \right)={{\text{e}}^{2x}}.\frac{1}{x}$ nên $\int\limits_{1}^{\ln 3}{xf\left( x \right)} \text{d}x=\int\limits_{1}^{\ln 3}{{{\text{e}}^{2x}}} \text{d}x=\left. \frac{1}{2}{{\text{e}}^{2x}} \right|_{1}^{\ln 3}=\frac{9}{2}-\frac{{{\text{e}}^{2}}}{2}=\frac{9-{{\text{e}}^{2}}}{2}$.
$\Rightarrow {f}'\left( x \right)={{\left( {{\text{e}}^{2x}}.\frac{1}{x} \right)}^{\prime }}\Rightarrow f\left( x \right)={{\text{e}}^{2x}}.\frac{1}{x}+C$.
+ Do $f\left( 1 \right)={{\text{e}}^{2}}$ nên ${{\text{e}}^{2.1}}.\frac{1}{1}+C={{\text{e}}^{2}}\Leftrightarrow C=0$.
+ Vậy $f\left( x \right)={{\text{e}}^{2x}}.\frac{1}{x}$ nên $\int\limits_{1}^{\ln 3}{xf\left( x \right)} \text{d}x=\int\limits_{1}^{\ln 3}{{{\text{e}}^{2x}}} \text{d}x=\left. \frac{1}{2}{{\text{e}}^{2x}} \right|_{1}^{\ln 3}=\frac{9}{2}-\frac{{{\text{e}}^{2}}}{2}=\frac{9-{{\text{e}}^{2}}}{2}$.
Đáp án D.