$$
\int\frac{dx}{x\sqrt{x^2+1}} = \int\frac{x\,dx}{x^2\sqrt{x^2+1}} = \int\frac{1}{x^2\sqrt{x^2+1}}
\Big(x\,dx\Big)
$$
The big parentheses are of course a hint that what's inside them is to become $du$, or a constant times $du$. But should $u$ be $x^2$ or $x^2+1$? Either way, $\displaystyle\Big(x\,dx\Big)$ becomes $\displaystyle\Big( \frac12\,du\Big)$. I think usually it's better to have the thing under the radical be simple, so I'll say $u=x^2+1$, and we have
$$
\frac12\int\frac{du}{(u-1)\sqrt{u}}.
$$
We can rationalize $\sqrt{u}$ by letting
\begin{align}
w & = \sqrt{u} \\
w^2 & = u \\
2w\,dw & = du
\end{align}
and we have
$$
\frac12\int\frac{2w\,dw}{(w^2-1)w} = \int\frac{dw}{w^2-1}.
$$
Then use partial fractions, getting
$$
\int\left(\frac{A}{w-1}+\frac{B}{w+1}\right)\,dw
$$
and you need to figure out what $A$ and $B$ are.
That works, but a trigonometric substitution also comes to mind. The expression $\sqrt{x^2+1}$ should remind you of $\sqrt{\tan^2\theta+1}= \pm\sec\theta$, and if it doesn't remind you of that, that's something to work on. Review some trigonometry and trigonometric substitutions. If $x=\tan\theta$ then $dx=\sec^2\theta\,d\theta$, and we have
$$
\int\frac{\sec^2\theta\,d\theta}{\tan\theta\sec\theta} = \int\frac{\sec\theta\,d\theta}{\tan\theta} = \int\csc\theta\,d\theta.
$$
That's a hard one to do from scratch, but it's also one that you can look up in standard tables.