Integral of $\frac{\sqrt{x^2+1}}{x^2}$ How to integrate: $$\int \frac{\sqrt{x^2+1}}{x^2}\text dx$$
I've typed this into wolframalpha as a result I received something with inverse hyperbolic sine. Is it possible to have it without this hyperbolic sine? If so, how to do that?

Let $$ I= \displaystyle \int \frac{\sqrt{x^2+1}}{x^2}dx = \int \frac{\sqrt{x^2+1}}{x^2} \times \frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}dx $$
$$\displaystyle I  = \int\frac{x^2+1}{x^2\sqrt{x^2+1}}dx = \int\frac{1}{\sqrt{x^2+1}}dx+\int \frac{1}{x^2\sqrt{x^2+1}}dx$$
Now in second Integral , Put $\displaystyle x = \frac{1}{t}\;,$ Then $\displaystyle dx = -\frac{1}{t^2}dt$
So we get   $$\displaystyle I  = \ln\left|x+\sqrt{x^2+1}\right|+J$$
So Integral $J$ Convert into $$\displaystyle -\int \frac{t}{\sqrt{t^2+1}}dt\;,$$ Now Substitute $t^2+1=u^2\;,$ Then $tdt = udu$
So  Second Integral $J$ into $$\displaystyle -\int \frac{udu}{u} = -u=-\sqrt{t^2+1}=-\frac{\sqrt{x^2+1}}{x}$$
So $$\displaystyle I = \frac{\sqrt{x^2+1}}{x^2}dx= \ln\left|x+\sqrt{x^2+1}\right|-\frac{\sqrt{x^2+1}}{x}+\mathcal{C}$$
