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Câu hỏi: Cho $\int\limits_{e}^{{{e}^{2}}}{\dfrac{3dx}{x\left( \sqrt{\ln x+1}+\sqrt{\ln x} \right)}}=a\sqrt{3}-b\sqrt{2}+c$, trong đó a, b, c là các số nguyên dương. Giá trị của biểu thức $T=a+b+c$ bằng
A. 16
B. 13
C. 11
D. 15
A. 16
B. 13
C. 11
D. 15
Ta có: $\int\limits_{e}^{{{e}^{2}}}{\dfrac{3dx}{x\left( \sqrt{\ln x+1}+\sqrt{\ln x} \right)}}=\int\limits_{e}^{{{e}^{2}}}{\dfrac{3\left( \sqrt{\ln x+1}-\sqrt{\ln x} \right)dx}{x\left( \sqrt{\ln x+1}+\sqrt{\ln x} \right)\left( \sqrt{\ln x+1}-\sqrt{\ln x} \right)}}$
$=\int\limits_{e}^{{{e}^{2}}}{\dfrac{3\left( \sqrt{\ln x+1}-\sqrt{\ln x} \right)}{x}dx}=\int\limits_{e}^{{{e}^{2}}}{3\left( \sqrt{\ln x+1}-\sqrt{\ln x} \right)d\left( \ln x \right)}=\left. \left( 2\sqrt{{{\left( \ln x+1 \right)}^{3}}}-2\sqrt{{{\ln }^{3}}x} \right) \right|_{e}^{{{e}^{2}}}$
$=2\left( \sqrt{27}-\sqrt{8}-\sqrt{8}+1 \right)=6\sqrt{3}-8\sqrt{2}+2$. Suy ra $a=6,b=8,c=2\Rightarrow a+b+c=16$.
$=\int\limits_{e}^{{{e}^{2}}}{\dfrac{3\left( \sqrt{\ln x+1}-\sqrt{\ln x} \right)}{x}dx}=\int\limits_{e}^{{{e}^{2}}}{3\left( \sqrt{\ln x+1}-\sqrt{\ln x} \right)d\left( \ln x \right)}=\left. \left( 2\sqrt{{{\left( \ln x+1 \right)}^{3}}}-2\sqrt{{{\ln }^{3}}x} \right) \right|_{e}^{{{e}^{2}}}$
$=2\left( \sqrt{27}-\sqrt{8}-\sqrt{8}+1 \right)=6\sqrt{3}-8\sqrt{2}+2$. Suy ra $a=6,b=8,c=2\Rightarrow a+b+c=16$.
Đáp án A.