Cho

B

- Hướng dẫn giải

Đáp án B

Ta có:

$$\begin{gathered} \underset{\frac{1}{3}}{\overset{1}{\int }}\frac{x}{3x+\sqrt{9x^2-1}}dx \\ =\underset{\frac{1}{3}}{\overset{1}{\int }}\frac{x\left(3x-\sqrt{9x^2-1}\right)}{9x^2-9x^2+1}dx \\ =\underset{\frac{1}{3}}{\overset{1}{\int }}\left(3x^2-x\sqrt{9x^2-1}\right)dx \end{gathered}$$

 $$\begin{gathered} =\underset{\frac{1}{3}}{\overset{1}{\int }}3x^{2}dx-\frac{1}{18}\underset{\frac{1}{3}}{\overset{1}{\int }}\sqrt{9x^2-1}d\left(9x^2-1\right)= \\ {\left.\left(x^3-\frac{1}{18}.\frac{2}{3}\sqrt{{\left(9x^2-1\right)}^3}\right)\right|}_{\frac{1}{3}}^1 \\ =\frac{26}{27}-\frac{16}{27}\sqrt{2} \end{gathered}$$

Suy ra $a=\frac{26}{27}; b=\frac{-16}{27}$