Câu hỏi: Cho $\int\limits_{16}^{55}{\dfrac{dx}{x\sqrt{x+9}}=a\ln 2+b\ln 5+c\ln 11}$ với $a,b,c$ là các số hữu tỉ. Mệnh đề nào dưới đây đúng?
A. $a-b=-c.$
B. $a+b=c.$
C. $a+b=3c.$
D. $a-b=-3c.$
A. $a-b=-c.$
B. $a+b=c.$
C. $a+b=3c.$
D. $a-b=-3c.$
Đặt $t=\sqrt{x+9}\Rightarrow {{t}^{2}}=x+9\Rightarrow \left[ \begin{aligned}
& 2tdt=dx \\
& x={{t}^{2}}-9 \\
\end{aligned} \right.$
Đổi cận $\left[ \begin{aligned}
& x=16\Rightarrow t=5 \\
& x=55\Rightarrow t=8 \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{16}^{55}{\dfrac{dx}{x\sqrt{x+9}}}=\int\limits_{5}^{8}{\dfrac{2tdt}{\left( {{t}^{2}}-9 \right)t}}=2\int\limits_{5}^{8}{\dfrac{dt}{{{t}^{2}}-9}}=\dfrac{1}{3}\int\limits_{5}^{8}{\left( \dfrac{1}{t-3}-\dfrac{1}{t+3} \right)dt}=\dfrac{1}{3}\ln \left| t-3 \right|\left| \begin{aligned}
& 8 \\
& 5 \\
\end{aligned} \right.-\dfrac{1}{3}\ln \left| t+3 \right|\left| \begin{aligned}
& 8 \\
& 5 \\
\end{aligned} \right.$
$=\dfrac{2}{3}\ln 2+\dfrac{1}{3}\ln 5-\dfrac{1}{3}\ln 11\Rightarrow \left\{ \begin{aligned}
& a=\dfrac{2}{3} \\
& b=\dfrac{1}{3} \\
& c=-\dfrac{1}{3} \\
\end{aligned} \right.\Rightarrow a-b=-c.$
& 2tdt=dx \\
& x={{t}^{2}}-9 \\
\end{aligned} \right.$
Đổi cận $\left[ \begin{aligned}
& x=16\Rightarrow t=5 \\
& x=55\Rightarrow t=8 \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{16}^{55}{\dfrac{dx}{x\sqrt{x+9}}}=\int\limits_{5}^{8}{\dfrac{2tdt}{\left( {{t}^{2}}-9 \right)t}}=2\int\limits_{5}^{8}{\dfrac{dt}{{{t}^{2}}-9}}=\dfrac{1}{3}\int\limits_{5}^{8}{\left( \dfrac{1}{t-3}-\dfrac{1}{t+3} \right)dt}=\dfrac{1}{3}\ln \left| t-3 \right|\left| \begin{aligned}
& 8 \\
& 5 \\
\end{aligned} \right.-\dfrac{1}{3}\ln \left| t+3 \right|\left| \begin{aligned}
& 8 \\
& 5 \\
\end{aligned} \right.$
$=\dfrac{2}{3}\ln 2+\dfrac{1}{3}\ln 5-\dfrac{1}{3}\ln 11\Rightarrow \left\{ \begin{aligned}
& a=\dfrac{2}{3} \\
& b=\dfrac{1}{3} \\
& c=-\dfrac{1}{3} \\
\end{aligned} \right.\Rightarrow a-b=-c.$
Đáp án A.