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Câu hỏi: Cho $\int_{\dfrac{1}{3}}^1 \dfrac{x}{3 x+\sqrt{9 x^2-1}} \mathrm{~d} x=a+b \sqrt{2}$, với $a, b$ là các số hữu tì. Khi đó giá trị của $a$ là
A. $-\dfrac{26}{27}$.
B. $\dfrac{26}{27}$.
C. $-\dfrac{27}{26}$.
D. $-\dfrac{25}{27}$.
A. $-\dfrac{26}{27}$.
B. $\dfrac{26}{27}$.
C. $-\dfrac{27}{26}$.
D. $-\dfrac{25}{27}$.
$
\begin{aligned}
& I=\int_{\dfrac{1}{3}}^1 \dfrac{x}{3 x+\sqrt{9 x^2-1}} \mathrm{~d} x=\int_{\dfrac{1}{3}}^1 \dfrac{x\left(3 x-\sqrt{9 x^2-1}\right)}{9 x^2-\left(9 x^2-1\right)} \mathrm{d} x=\int_{\dfrac{1}{3}}^1\left(3 x^2-x \sqrt{9 x^2-1}\right) \mathrm{d} x . \\
& =\int_{\dfrac{1}{3}}^1 3 x^2 \mathrm{~d} x-\int_{\dfrac{1}{3}}^1 x \sqrt{9 x^2-1} \mathrm{~d} x=\left.x^3\right|_{\dfrac{1}{3}} ^1-J=\dfrac{26}{27}-J . \\
& \text { Xét } J=\int_{\dfrac{1}{3}}^1 x \sqrt{9 x^2-1} \mathrm{~d} x .
\end{aligned}
$
Đặt $t=\sqrt{9 x^2-1} \Rightarrow \mathrm{d} t=\dfrac{9 x}{\sqrt{9 x^2-1}} \mathrm{~d} x \Rightarrow x \mathrm{~d} x=\dfrac{1}{9} t \mathrm{~d} t$.
Đổi cận: $x=1 \Rightarrow t=2 \sqrt{2} ; x=\dfrac{1}{3} \Rightarrow t=0$.
$
\Rightarrow J=\dfrac{1}{9} \int_0^{2 \sqrt{2}} t^2 \mathrm{~d} t=\left.\dfrac{1}{27} t^3\right|_0 ^{2 \sqrt{2}}=\dfrac{16 \sqrt{2}}{27} \Rightarrow I=\dfrac{26}{27}-\dfrac{16 \sqrt{2}}{27} \Rightarrow a=\dfrac{26}{27} .
$
\begin{aligned}
& I=\int_{\dfrac{1}{3}}^1 \dfrac{x}{3 x+\sqrt{9 x^2-1}} \mathrm{~d} x=\int_{\dfrac{1}{3}}^1 \dfrac{x\left(3 x-\sqrt{9 x^2-1}\right)}{9 x^2-\left(9 x^2-1\right)} \mathrm{d} x=\int_{\dfrac{1}{3}}^1\left(3 x^2-x \sqrt{9 x^2-1}\right) \mathrm{d} x . \\
& =\int_{\dfrac{1}{3}}^1 3 x^2 \mathrm{~d} x-\int_{\dfrac{1}{3}}^1 x \sqrt{9 x^2-1} \mathrm{~d} x=\left.x^3\right|_{\dfrac{1}{3}} ^1-J=\dfrac{26}{27}-J . \\
& \text { Xét } J=\int_{\dfrac{1}{3}}^1 x \sqrt{9 x^2-1} \mathrm{~d} x .
\end{aligned}
$
Đặt $t=\sqrt{9 x^2-1} \Rightarrow \mathrm{d} t=\dfrac{9 x}{\sqrt{9 x^2-1}} \mathrm{~d} x \Rightarrow x \mathrm{~d} x=\dfrac{1}{9} t \mathrm{~d} t$.
Đổi cận: $x=1 \Rightarrow t=2 \sqrt{2} ; x=\dfrac{1}{3} \Rightarrow t=0$.
$
\Rightarrow J=\dfrac{1}{9} \int_0^{2 \sqrt{2}} t^2 \mathrm{~d} t=\left.\dfrac{1}{27} t^3\right|_0 ^{2 \sqrt{2}}=\dfrac{16 \sqrt{2}}{27} \Rightarrow I=\dfrac{26}{27}-\dfrac{16 \sqrt{2}}{27} \Rightarrow a=\dfrac{26}{27} .
$
Đáp án B.