Ta có: \(I=-\int_{1}^{\sqrt{3}}ln(x^2+1).d(\frac{1}{x})=-\frac{1}{x}.ln(x^2+1)\bigg |_{1}^{\sqrt{3}}+\int_{1}^{\sqrt{3}}\frac{1}{x}dln(x^2+1)\) \(=\frac{3-2\sqrt{3}}{3}ln2+2\int_{1}^{\sqrt{3}}\frac{dx}{x^2+1}\) Tính \(\int_{1}^{\sqrt{3}}\frac{dx}{x^2+1}\). Đặt x = tant, khi đó \(dx=\frac{dt}{cos^2t}\), với x = 1 thì \(t=\frac{\pi }{4}\) với \(x=\sqrt{3}\) thì \(t=\frac{\pi }{3}\) Suy ra \(\int_{1}^{\sqrt{3}}\frac{dx}{x^2+1}=\int_{\frac{\pi }{4}}^{\frac{\pi }{3}}dt=\frac{\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}\) \(\Rightarrow 2\int_{1}^{\sqrt{3}}\frac{dx}{x^2+1}=\frac{\pi }{6}\) Vậy \(I=\frac{3-2\sqrt{3}}{3}ln2+\frac{\pi }{6}\)