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