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