A long trigonometric/hyperbolic integral Evaluate:
$$ \int \frac {dx}{\sqrt{1-x^2}+\sqrt{x^2+1}}=\int \frac{\sqrt{1-x^2}-\sqrt{x^2+1}}{-2x^2}dx=-\frac{1}{2}\left(\int \frac{\sqrt{1-x^2}}{x^2}dx-\int\frac{\sqrt{x^2+1}}{x^2}dx\right) $$
In the first integral let $x=\sin\theta$, $dx=\cos\theta d\theta$
In the second integral let $x=\sinh \varphi$, $dx=\cosh\varphi d\varphi$
$$-\frac{1}{2}\left(\int \frac{\sqrt{1-\sin^2\theta}}{\sin^2\theta}\cos\theta d\theta-\int\frac{\sqrt{\sinh^2\varphi+1}}{\sinh^2\varphi}\cosh\varphi d\varphi\right)=-\frac{1}{2}\left(\int \frac{\cos^2\theta}{\sin^2\theta}d\theta-\int\frac{{\cosh^2\varphi}}{\sinh^2\varphi}d\varphi\right)=-\frac{1}{2}\left(\int \frac{1-\sin^2\theta}{\sin^2\theta}d\theta-\int\frac{{1+\sin^2\varphi}}{\sinh^2\varphi}d\varphi\right)=-\frac{1}{2}\left(-\cot\theta -\theta+\coth\varphi-\varphi\right)+c=\frac{1}{2}\left(\cot(\arcsin x)+\arcsin x-\coth(arcsinh x)+arcsinh x\right)+c$$
I'm not sure if it is right. Thanks in advance.
 A: $$I = \int \frac {dx}{\sqrt{1-x^2}+\sqrt{x^2+1}}$$
$$\int \frac{\sqrt{1-x^2}-\sqrt{x^2+1}}{-2x^2}dx$$
$$-\frac{1}{2}\left(\int \frac{\sqrt{1-x^2}}{x^2}dx-\int\frac{\sqrt{x^2+1}}{x^2}dx\right) $$
Apply integration By Parts:
$$I_1 = \int \frac{\sqrt{1-x^2}}{x^2}dx$$
$$\sqrt{1 - x^2}\left( -\frac{1}{x}\right) -  \int -\frac{x}{\sqrt{1-x^2}}\left( -\frac{1}{x}\right)dx$$
$$-\frac{\sqrt{1 - x^2}}{x} -  \int \frac{1}{\sqrt{1-x^2}}dx$$
$$I_1 = -\frac{\sqrt{1 - x^2}}{x} -  \arcsin(x) +C$$
Apply integration By Parts:
$$I_2 = \int\frac{\sqrt{x^2+1}}{x^2}dx$$
$$\int\frac{\sqrt{x^2+1}}{x^2}dx$$
$$\sqrt{x^2+1}\left( -\frac{1}{x}\right) -  \int \frac{x}{\sqrt{x^2+1}}\left( -\frac{1}{x}\right)dx$$
$$-\frac{\sqrt{x^2+1}}{x} +  \int \frac{1}{\sqrt{x^2+1}}dx$$
$$I_2-\frac{\sqrt{x^2+1}}{x} +  \operatorname{arcsch}(x) + C$$

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

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


And the solution coincides with WolframAlpha :D
$$\int \frac{dx}{\sqrt{1-x^2}+\sqrt{x^2+1}} dx = \frac{\sqrt{1 - x^2} - \sqrt{1 + x^2} + x \sin^{-1}(x) + x \sinh^{-1}(x))}{2 x} + C$$
