物理

在哪里讲过啊／／／／

$\color{red}{}\int\dfrac{\mathrm{d}x}{\sqrt{1+x^2}}=\int\dfrac{\mathrm{d}(\tanθ)}{\sqrt{1+\tan^2θ}}=\color{green}{}\int\dfrac{\mathrm{d}θ}{\cosθ}=\int\dfrac{\cosθ\mathrm{d}θ}{\cos^2θ}=\int\dfrac{\mathrm{d}(\sinθ)}{1-\sin^2θ}=\color{blue}{}\int\dfrac{\mathrm{d}y}{1-y^2}=\int\dfrac{\mathrm{d}y}{(1+y)(1-y)}=\dfrac{1}{2}\int(\dfrac{1}{1+y}+\dfrac{1}{1-y})\mathrm{d}y=\dfrac{1}{2}(\int\dfrac{\mathrm{d}(1+y)}{1+y}-\int\dfrac{\mathrm{d}(1-y)}{1-y})=\dfrac{1}{2}[\ln(1+y)-\ln(1-y)]\color{green}{}=\dfrac{1}{2}[\ln(1+\sinθ)-\ln(1-\sinθ)]=\ln\dfrac{1+\sinθ}{\cosθ}\color{red}{}=\ln(\sqrt{1+\tan^2θ}+\tanθ)=\ln(\sqrt{1+x^2}+x)$