Example 13. Find $\int \frac{d x}{x^{2} \sqrt{1+x^{2}}}$.
Solution. Let $x=\operatorname{tg} t, \sqrt{1+x^{2}}=\frac{1}{\cos t}, \frac{1}{x^{2}}=\frac{\cos ^{2} t}{\sin ^{2} t}, d x=$ $=\frac{d t}{\cos ^{2} t}, t=\operatorname{arctg} x$. We arrive at an "almost tabular" integral, which after reverse substitutions takes the form $\left(\sin t=\frac{\operatorname{tg} t}{\sqrt{1+\operatorname{tg}^{2} t}}=\right.$ $\left.=\frac{x}{\sqrt{1+x^{2}}}\right)$

$$
\int \frac{d x}{x^{2} \sqrt{1+x^{2}}}=\int \frac{\cos t d t}{\sin ^{2} t}=-\frac{1}{\sin t}+C=C-\frac{\sqrt{1+x^{2}}}{x}
$$