Cho biết

B

- Hướng dẫn giải

Đáp án B

Đặt

$u=\ln \left(9-x^2\right)\Rightarrow du=\frac{-2x}{9-x^2}$

$dv=dx\Rightarrow v=x$

$I={\left.x.\ln \left(9-x^2\right)\right|}_0^1-\underset{0}{\overset{1}{\int }}\frac{-2x^2}{9-x^2}dx$

$={\left.x.\ln \left(9-x^2\right)\right|}_0^1+\underset{0}{\overset{1}{\int }}\left(-2+\frac{18}{9-x^2}\right)dx$

$=\ln 8-2+{\left.3\ln \left|\frac{3+x}{3-x}\right|\right|}_0^1=\ln 8-2+3\ln 2=6ln2-2$

Vậy $S=\left|6-2\right|=4.$