Ta có \(I=\int_{\sqrt{2}}^{\sqrt{5}}(2x+\sqrt{x^2-1})dx=\int_{\sqrt{2}}^{\sqrt{5}}2xdx+\int_{\sqrt{2}}^{\sqrt{5}}\sqrt{x^2-1}dx\) Tính \(I_1=\int_{\sqrt{2}}^{\sqrt{5}}2xdx=x^2\bigg|^{\sqrt{5}}_{\sqrt{2}}=5-2=3\) Tính \(I_2=\int_{\sqrt{2}}^{\sqrt{5}}\sqrt{x^2-1}dx\) Đặt \(\left\{\begin{matrix} u=\sqrt{x^2-1}\\ dv=dx \end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{x}{\sqrt{x^2-1}}dx\\ v=x \end{matrix}\right.\). Khi đó \(I_2=x\sqrt{x^2-1} \bigg|^{\sqrt{5}}_{\sqrt{2}}-\int_{\sqrt{2}}^{\sqrt{5}}\frac{x^2}{\sqrt{x^2-1}}dx=2\sqrt{5}-\sqrt{2}-\int_{\sqrt{2}}^{\sqrt{5}}\frac{x^2}{\sqrt{x^2-1}}\) \(=2\sqrt{5}-\sqrt{2}-\int_{\sqrt{2}}^{\sqrt{5}}\frac{x^2-1+1}{\sqrt{x^2-1}}dx\) \(=2\sqrt{5}-\sqrt{2}-\int_{\sqrt{2}}^{\sqrt{5}}\sqrt{x^2-1}dx-\int_{\sqrt{2}}^{\sqrt{5}}\frac{dx}{\sqrt{x^2-1}}\) Suy ra \(2I_2=(2\sqrt{5}-\sqrt{2})-\int_{\sqrt{2}}^{\sqrt{5}}\frac{dx}{\sqrt{x^2-1}} =(2\sqrt{5}-\sqrt{2})-ln\left | (x+\sqrt{x^2-1}) \right |\bigg|^{\sqrt{5}}_{\sqrt{2}}\)\(\Rightarrow I_2=\frac{1}{2}(2\sqrt{5}-\sqrt{2})-\frac{1}{2}ln\frac{\sqrt{5}+2}{\sqrt{2}+1}\) Vậy \(I=3+\frac{1}{2}(2\sqrt{5}-\sqrt{2})-\frac{1}{2}ln\frac{\sqrt{5}+2}{\sqrt{2}+1}\)