Cho \(I=\int\limits^{\sqrt{3}}_1\frac{\sqrt{1+x^2}dx}{x^2}\), đặt \(\dfrac{\sqrt{1+x^2}}{x}=t\). Khi đó, $I=$
\(\int\limits^{\frac{2}{\sqrt{3}}}_{\sqrt{2}}\frac{t\text{d}t}{t^2-1}\). \(\int\limits^3_2\frac{t\text{d}t}{t^2+1}\). \(-\int\limits^{\frac{2}{\sqrt{3}}}_{\sqrt{2}}\frac{t\text{d}t}{t^2-1}\). \(\int\limits^3_2\frac{t\text{d}t}{t^2+1}\). Hướng dẫn giải:Đặt \(\dfrac{\sqrt{1+x^2}}{x}=t\Rightarrow\dfrac{1+x^2}{x^2}=t^2\Rightarrow\dfrac{1}{x^2}=t^2-1\) \(\Rightarrow x^2=\frac{1}{t^2-1}\Rightarrow2x\text{d}x=\frac{-2t\text{d}t}{\left(t^2-1\right)^2}\) . Do đó
\(\frac{\sqrt{1+x^2}}{x^2}\text{d}x=t.\frac{x\text{d}x}{x^2}=-t.\frac{\frac{t\text{d}t}{\left(t^2-1\right)^2}}{\frac{1}{t^2-1}}=-\frac{t^2\text{d}t}{t^2-1}\) và \(I=-\int\limits^{\frac{2}{\sqrt{3}}}_{\sqrt{2}}\frac{tdt}{t^2-1}\) .