Cho \(I=\int\limits^2_1\frac{x\text{d}x}{1+\sqrt{x-1}}\), đặt \(t=\sqrt{x-1}\). Khi đó, $I=$
\(\int\limits^1_0\frac{t^2+1}{t+1}\text{d}t\). \(\int\limits^1_0\frac{2t^3+t}{t+1}\text{d}t\). \(\int\limits^1_0\left(2t^2+2t+4-\frac{4}{t+1}\right)\text{d}t\). \(\int\limits^1_0\left(2t^2-2t+4-\frac{4}{t+1}\right)\text{d}t\). Hướng dẫn giải:Ta có: \(t=\sqrt{x-1}\Rightarrow x=t^2+1;\text{d}x=2t\text{d}t\)
Đổi cận: \(x|^2_1\Rightarrow t|^1_0\)
Suy ra:
\(I=\int\limits^1_0\frac{t^2+1}{1+t}2t\text{d}t\)
\(=\int\limits^1_0\frac{2t^3+2t}{t+1}\text{d}t\)
\(=\int\limits^1_0\frac{2t^3+2t^2-2t^2-2t+4t+4-4}{t+1}\text{d}t\)
\(=\int\limits^1_0\frac{2t^2\left(t+1\right)-2t\left(t+1\right)+4\left(t+1\right)-4}{t+1}\text{d}t\)
\(=\int\limits^1_0\left(2t^2-2t+4-\frac{4}{t+1}\right)\text{d}t\).