A long trigonometric/hyperbolic integral

Question

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.

Answer 1

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