直接積就完事了：
\begin{aligned} &{}\int{4\arctan{\frac{\sqrt{a^2-x^2}}{\rho-x}}\,\mathrm{d}x}\\ =&{}4x\arctan{\frac{\sqrt{a^2-x^2}}{\rho-x}}-\int{4x\,\mathrm{d}\arctan\frac{\sqrt{a^2-x^2}}{\rho-x}}\\ =&{}4x\arctan{\frac{\sqrt{a^2-x^2}}{\rho-x}}-\int{4x\cdot\frac{1}{1+\Big(\frac{\sqrt{a^2-x^2}}{\rho-x}\Big)^2}\cdot\frac{(\rho-x)\cdot\frac{-2x}{2\sqrt{a^2-x^2}}-(-1)\cdot\sqrt{a^2-x^2}}{(\rho-x)^2}\,\mathrm{d}x}\\ =&{}4x\arctan{\frac{\sqrt{a^2-x^2}}{\rho-x}}-\int{4x\cdot\frac{1}{(\rho-x)^2+a^2-x^2}\cdot\frac{-x(\rho-x)+a^2-x^2}{\sqrt{a^2-x^2}}\,\mathrm{d}x}\\ \xlongequal{x=a\sin\theta}&{}4x\arctan{\frac{\sqrt{a^2-x^2}}{\rho-x}}-\int{4a\sin\theta\cdot\frac{1}{\rho^2+a^2-2\rho a\sin\theta}\cdot\frac{a^2-\rho a\sin\theta}{a\cos\theta}a\cos\theta\,\mathrm{d}\theta}\\ \xlongequal{\lambda=\frac{\rho}{a}}&{}4x\arctan{\frac{\sqrt{a^2-x^2}}{\rho-x}}-4a\int{\frac{\sin\theta(1-\lambda\sin\theta)}{\lambda^2+1-2\lambda\sin\theta}\,\mathrm{d}\theta}\\ \xlongequal[t=\tan\frac{\theta}{2}]{\text{萬能公式}}&{}4x\arctan{\frac{\sqrt{a^2-x^2}}{\rho-x}}-4a\left(\Big(\frac{\lambda}{4}-\frac{1}{4 \lambda }\Big)\theta -\frac{\cos \theta }{2}+\Big(\frac{\lambda}{2} +\frac{1}{2\lambda} \Big) \arctan \frac{1-\lambda t}{\lambda -t}\right)+C\\ =&{}4x\arctan{\frac{\sqrt{a^2-x^2}}{\rho-x}}+\left(\frac{a^2 }{\rho } - \rho\right)\theta +2 a \cos \theta-\left(\frac{2 a^2 }{\rho }+2 \rho\right)\arctan\frac{a -\rho t}{\rho -at}+C \end{aligned}
x 從 -a 到 a ，相當於 \theta 從 -\frac{\pi}{2} 到 \frac{\pi}{2} ， t 從 -1 到 1 。直接計算得
\begin{aligned} &{}\int_{-a}^a{4\arctan\frac{\sqrt{a^2-x^2}}{\rho-x}\,\mathrm{d}x}\\ =&{}\left(0+\Big(\frac{a^2}{\rho}-\rho\Big)\frac{\pi}{2}+0-\Big(\frac{2a^2}{\rho}+2\rho\Big)\arctan(-1)\right)\\ &{}-\left(0-\Big(\frac{a^2}{\rho}-\rho\Big)\frac{\pi}{2}+0-\Big(\frac{2a^2}{\rho}+2\rho\Big)\arctan1\right)\\ =&{}\frac{a^2\pi}{\rho}-\left(-\frac{a^2\pi}{\rho}\right)\\ =&{}\frac{2a^2\pi}{\rho} \end{aligned}
