bài 35 tương tự [tex]t=\sqrt{x+9}[/tex]
=>tdt=dx và[tex]x=t^2-9[/tex] thay vao ta được [tex]\int _5^8 \frac{tdt}{\left ( t-3 \right )\left ( t+3 \right )t}dt =\frac{1}{6}\int _5^8 \frac{t+3-\left ( t-3 \right )}{\left ( t+3 \right )\left ( t-3 \right )}dt =\frac{1}{6}\int _5^8 \left (\frac{1}{t-3}- \frac{1}{t+3}\right )dt =\frac{1}{6}ln\left ( \frac{t-3}{t+3} \right )\int _5^8 =\frac{1}{6}\left ( 2ln2+ln5-ln11 \right )[/tex]
vậy đáp án A