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Câu hỏi: Cho $\int\limits_{8}^{56}{\dfrac{d\text{x}}{(x-1)\sqrt{x+8}}}=a\ln 5+b\ln 7+c\ln 11$, với a, b, c là các số hữu tỷ. Đặt $T=a+b-3c$ thì
A. $T\in (-1;0)$
B. $T\in (0;1)$
C. $T\in (1;2)$
D. $T\in (2;4)$
A. $T\in (-1;0)$
B. $T\in (0;1)$
C. $T\in (1;2)$
D. $T\in (2;4)$
Đặt $t=\sqrt{x+8}\Rightarrow {{t}^{2}}=x+8\Rightarrow 2t\text{d}t=d\text{x}$. Đổi cận $\left| \begin{aligned}
& x=8\Rightarrow t=4 \\
& x=56\Rightarrow t=8 \\
\end{aligned} \right.$. Khi đó
$\int\limits_{8}^{56}{\dfrac{dx}{(x-1)\sqrt{x+8}}}=\int\limits_{4}^{8}{\dfrac{2t\text{d}t}{({{t}^{2}}-9)t}}=\int\limits_{4}^{8}{\dfrac{2dt}{({{t}^{2}}-9)}}=\left. \dfrac{1}{3}\ln \left| \dfrac{t-3}{t+3} \right| \right|_{4}^{8}=\dfrac{1}{3}\ln \dfrac{35}{11}=\dfrac{1}{3}\ln 5+\dfrac{1}{3}\ln 5-\dfrac{1}{3}\ln 11$ nên
$a=\dfrac{1}{3},b=\dfrac{1}{3},c=-\dfrac{1}{3}\Rightarrow a+b-3c=\dfrac{5}{3}\in (1;2)$.
& x=8\Rightarrow t=4 \\
& x=56\Rightarrow t=8 \\
\end{aligned} \right.$. Khi đó
$\int\limits_{8}^{56}{\dfrac{dx}{(x-1)\sqrt{x+8}}}=\int\limits_{4}^{8}{\dfrac{2t\text{d}t}{({{t}^{2}}-9)t}}=\int\limits_{4}^{8}{\dfrac{2dt}{({{t}^{2}}-9)}}=\left. \dfrac{1}{3}\ln \left| \dfrac{t-3}{t+3} \right| \right|_{4}^{8}=\dfrac{1}{3}\ln \dfrac{35}{11}=\dfrac{1}{3}\ln 5+\dfrac{1}{3}\ln 5-\dfrac{1}{3}\ln 11$ nên
$a=\dfrac{1}{3},b=\dfrac{1}{3},c=-\dfrac{1}{3}\Rightarrow a+b-3c=\dfrac{5}{3}\in (1;2)$.
Đáp án C.