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Câu hỏi: Biết $\int\limits_{1}^{e}{\dfrac{\ln x}{x\sqrt{1+\ln x}}\text{d}x}=a+b\sqrt{2}$, với $a;b\in \mathbb{Q}$. Tính $a-b$.
A. $\dfrac{2}{3}$.
B. 2.
C. $e-2$.
D. $3$.
A. $\dfrac{2}{3}$.
B. 2.
C. $e-2$.
D. $3$.
Đặt $t=\sqrt{1+\ln x}\Rightarrow {{t}^{2}}=1+\ln x\Rightarrow 2t.dt=\dfrac{1}{x}.dx$
Đổi cận: $\begin{matrix}
x=e & \Rightarrow & t=\sqrt{2} \\
x=1 & \Rightarrow & t=1 \\
\end{matrix}$
Khi đó: $\int\limits_{1}^{e}{\dfrac{\ln x}{x\sqrt{1+\ln x}}\text{d}x}=\int\limits_{1}^{\sqrt{2}}{\dfrac{{{t}^{2}}-1}{t}\text{2}t.\text{d}}t=\int\limits_{1}^{\sqrt{2}}{\left( 2{{t}^{2}}-2 \right)\text{d}}t=\left. \left( \dfrac{2{{t}^{3}}}{3}-2t \right) \right|_{1}^{\sqrt{2}}$
$=\dfrac{2.2\sqrt{2}}{3}-2\sqrt{2}-\left( \dfrac{2}{3}-2 \right)=\dfrac{4}{3}-\dfrac{2}{3}\sqrt{2}$
Vậy $a=\dfrac{4}{3}; b=-\dfrac{2}{3}\Rightarrow a-b=2$.
Đổi cận: $\begin{matrix}
x=e & \Rightarrow & t=\sqrt{2} \\
x=1 & \Rightarrow & t=1 \\
\end{matrix}$
Khi đó: $\int\limits_{1}^{e}{\dfrac{\ln x}{x\sqrt{1+\ln x}}\text{d}x}=\int\limits_{1}^{\sqrt{2}}{\dfrac{{{t}^{2}}-1}{t}\text{2}t.\text{d}}t=\int\limits_{1}^{\sqrt{2}}{\left( 2{{t}^{2}}-2 \right)\text{d}}t=\left. \left( \dfrac{2{{t}^{3}}}{3}-2t \right) \right|_{1}^{\sqrt{2}}$
$=\dfrac{2.2\sqrt{2}}{3}-2\sqrt{2}-\left( \dfrac{2}{3}-2 \right)=\dfrac{4}{3}-\dfrac{2}{3}\sqrt{2}$
Vậy $a=\dfrac{4}{3}; b=-\dfrac{2}{3}\Rightarrow a-b=2$.
Đáp án B.