Cho \(I=\int\limits^8_3\frac{xdx}{1+\sqrt{x+1}}\), đặt \(t=\sqrt{x+1}\) thì $I$ bằng
\(\int\limits^3_2\left(1-t^2\right)\text{d}t\). \(2\int\limits^3_2\left(t^2-t\right)\text{d}t\). \(2\int\limits^8_3\left(t-t^2\right)\text{d}t\). \(\int\limits^3_2\left(t+t^2\right)\text{d}t\). Hướng dẫn giải:Đặt \(t=\sqrt{x+1}\) suy ra \(x=t^2-1\)
\(\Rightarrow\text{d}x=2t\text{d}t\)
Đổi cận \(x|^8_3\Rightarrow t|^3_2\)
Vậy
\(I=\int\limits^3_2\frac{\left(t^2-1\right).2t\text{d}t}{1+t}\)
\(=\int\limits^3_2\frac{\left(t-1\right)\left(t+1\right)2t}{t+1}\text{d}t\)
\(=2\int\limits^3_2\left(t^2-t\right)\text{d}t\).
