一个长不定积分$\int \frac {dx}{\sqrt{1-x^2}+\sqrt{x^2+1}}$

提问：求解：

$$\int \frac {dx}{\sqrt{1-x^2}+\sqrt{x^2+1}}=\int \frac{\sqrt{1-x^2}-\sqrt{x^2+1}}{-2x^2}dx=-\frac{1}{2}\left(\int \frac{\sqrt{1-x^2}}{x^2}dx-\int\frac{\sqrt{x^2+1}}{x^2}dx\right)$$

在第一个积分式中令$x=\sin\theta$，$dx=\cos\theta d\theta$

在第二个积分式中令$x=\sinh \varphi$，$dx=\cosh\varphi d\varphi$

$$-\frac{1}{2}\left(\int \frac{\sqrt{1-\sin^2\theta}}{\sin^2\theta}\cos\theta d\theta-\int\frac{\sqrt{\sinh^2\varphi+1}}{\sinh^2\varphi}\cosh\varphi d\varphi\right)=-\frac{1}{2}\left(\int \frac{\cos^2\theta}{\sin^2\theta}d\theta-\int\frac{{\cosh^2\varphi}}{\sinh^2\varphi}d\varphi\right)=-\frac{1}{2}\left(\int \frac{1-\sin^2\theta}{\sin^2\theta}d\theta-\int\frac{{1+\sin^2\varphi}}{\sinh^2\varphi}d\varphi\right)=-\frac{1}{2}\left(-\cot\theta -\theta+\coth\varphi-\varphi\right)+c=\frac{1}{2}\left(\cot(\arcsin x)+\arcsin x-\coth(arcsinh x)+arcsinh x\right)+c$$

我不知道这是否是正确的，先提前感谢一下。

回答：$$I = \int \frac {dx}{\sqrt{1-x^2}+\sqrt{x^2+1}}$$

$$\int \frac{\sqrt{1-x^2}-\sqrt{x^2+1}}{-2x^2}dx$$

$$-\frac{1}{2}\left(\int \frac{\sqrt{1-x^2}}{x^2}dx-\int\frac{\sqrt{x^2+1}}{x^2}dx\right)$$

应用分部积分法：$$I_1 = \int \frac{\sqrt{1-x^2}}{x^2}dx$$

$$\int\frac{\sqrt{x^2+1}}{x^2}dx$$

$$\sqrt{x^2+1}\left( -\frac{1}{x}\right) - \int \frac{x}{\sqrt{x^2+1}}\left( -\frac{1}{x}\right)dx$$

$$-\frac{\sqrt{x^2+1}}{x} + \int \frac{1}{\sqrt{x^2+1}}dx$$

$$I_2-\frac{\sqrt{x^2+1}}{x} + \operatorname{arcsch}(x) + C$$

同时$$I = -\frac{1}{2}\left( \left(-\frac{\sqrt{1 - x^2}}{x} - \arcsin(x)\right) - \left( -\frac{\sqrt{x^2+1}}{x} + \operatorname{arcsch}(x) \right)\right) + C$$

$$I = -\frac{1}{2}\left( -\frac{\sqrt{1 - x^2}}{x} - \arcsin(x) + \frac{\sqrt{x^2+1}}{x} - \operatorname{arcsch}(x) \right) + C$$

并且答案跟 WolframAlpha的一致：

$$\int \frac{dx}{\sqrt{1-x^2}+\sqrt{x^2+1}} dx = \frac{\sqrt{1 - x^2} - \sqrt{1 + x^2} + x \sin^{-1}(x) + x \sinh^{-1}(x))}{2 x} + C$$

翻译自Mathstackexchange：A long integral $\int \frac {dx}{\sqrt{1-x^2}+\sqrt{x^2+1}}$