Đề thế này hả bạn: [tex]\int \frac{\sqrt{4-3x}}{x+1}dx[/tex]
Đặt [tex]\sqrt{4-3x}=t\Rightarrow x=\frac{4-t^2}{3}\Rightarrow dx=-\frac{2t}{3}dt[/tex]
[tex]I=\int \frac{-2}{3}.\frac{t^2}{\frac{4-t^2}{3}+1}dt[/tex] [tex]=\int \frac{2t^2}{t^2-7}dt=\int \left ( 2+\frac{14}{t^2-7} \right )dt=2t+\sqrt{7}.ln\left | \frac{t-\sqrt{7}}{t+\sqrt{7}} \right |+C[/tex]
[tex]=2\sqrt{4-3x}+\sqrt{7}.ln\left | \frac{\sqrt{4-3x}-\sqrt{7}}{\sqrt{4-3x}+\sqrt{7}} \right |+C[/tex]