int_(1)^sqrt(3) dfrac(d x)(x^2 sqrt(1+x^2))

解1$\int_{1}^{\sqrt{3}} \dfrac{d x}{x^{2} \sqrt{1+x^{2}}}=\int_{1}^{\sqrt{3}} \dfrac{d x}{x^{-3} \sqrt{1+x^{-2}}} $$=-\int_{1}^{\sqrt{3}} \dfrac{d\left(1+x^{-2}\right)}{2 \sqrt{1+x^{-2}}}=-\left.\sqrt{1+x^{-2}}\right|_{1} ^{\sqrt{3}} \\ $$=-\left(\sqrt{1+\dfrac{1}{3}}-\sqrt{2}\right)=\sqrt{2}-\sqrt{\dfrac{4}{3}}$$=\sqrt{2}-\dfrac{2 \sqrt{3}}{3}$解 2 令 $ x$=$\tan t $, 则当 $ x$=$1 $ 时取 $ t$=$\dfrac{\pi}{4} $, 当 $ x$=$\sqrt{3} $ 时取 $ t$=$\dfrac{\pi}{3}$,$ d x$=$\sec ^{2} t d t $, 进而$ \int_{1}^{\sqrt{3}} \dfrac{d x}{x^{2} \sqrt{1+x^{2}}}=\int_{\pi / 4}^{\pi / 3} \dfrac{\sec ^{2} t d t}{\tan ^{2} t \sqrt{1+\tan ^{2} t}}$$=\int_{\pi / 4}^{\pi / 3} \dfrac{\sec t d t}{\tan ^{2} t}=\int_{\pi / 4}^{\pi / 3} \dfrac{\cos t d t}{\sin ^{2} t}$$=\int_{\pi / 4}^{\pi / 3} \dfrac{d \sin t}{\sin ^{2} t}$$=\left.\left(-\dfrac{1}{\sin t}\right)\right|_{\pi / 4} ^{\pi / 3}=-\left(\dfrac{2}{\sqrt{3}}-\sqrt{2}\right)$$=\sqrt{2}-\dfrac{2 \sqrt{3}}{3}$