计算$$ \int_0^{\sqrt{3} a} \frac{d x}{d^2+x^2} $$

要计算$$ \int_0^{\sqrt{3}a} \frac{dx}{2\frac{d^2x}{dx^2}} $$

可以使用代换法，令$y=\frac{dx}{d\theta}$，则$\frac{d^2x}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{d\theta}\left(\frac{dy}{dx}\right)\frac{d\theta}{dx}=\frac{d}{d\theta}\left(\frac{dy}{dx}\right)\frac{1}{y}$，因此原积分可表示为

$$\int_0^{\sqrt{3}a} \frac{dx}{2\frac{d^2x}{dx^2}}=\int_0^{\sqrt{3}a} \frac{y d\theta}{2\frac{d}{d\theta}\left(\frac{dy}{dx}\right)\frac{1}{y}}=\int_0^{\sqrt{3}a} \frac{y^2}{2\frac{dy}{dx}\frac{d^2y}{dx^2}} d\theta$$

接下来可以使用分部积分法计算积分。令$u=y$，$dv=\frac{dy}{dx}\frac{d^2y}{dx^2}d\theta$，则$du=dy$，$v=\frac{1}{2}\left(\frac{dy}{dx}\right)^2$，于是有

$$\int_0^{\sqrt{3}a} \frac{y^2}{2\frac{dy}{dx}\frac{d^2y}{dx^2}} d\theta = \frac{y^2}{2}\left[\frac{1}{2}\left(\frac{dy}{dx}\right)^2\right]_0^{\sqrt{3}a}-\frac{1}{2}\int_0^{\sqrt{3}a} \left(\frac{dy}{dx}\right)^2 d\theta$$

注意到$\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{y}{dx/d\theta}$，因此

$$\left(\frac{dy}{dx}\right)^2=\left(\frac{y}{dx/d\theta}\right)^2=\left(\frac{y}{\frac{dx}{d\theta}}\right)^2=y^2$$

代入上式得

$$\int_0^{\sqrt{3}a} \frac{y^2}{2\frac{dy}{dx}\frac{d^2y}{dx^2}} d\theta = \frac{y^2}{4}\Big|_0^{\sqrt{3}a} -\frac{1}{2}\int_0^{\sqrt{3}a} y^2 d\theta=\frac{3a^2}{4}-\frac{a^2}{2}=\frac{a^2}{4}$$

因此，$$\int_0^{\sqrt{3}a} \frac{dx}{2\frac{d^2x}{dx^2}}=\frac{a^2}{4}$$