Answer
See proof
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{x^2}\sqrt {1 + {x^2}} }}} dx = - \frac{{\sqrt {1 + {x^2}} }}{x} + C \cr
& {\text{Let }}F\left( x \right) = - \frac{{\sqrt {1 + {x^2}} }}{x} + C\cr
&{\text{ We will show that F is an antiderivative of }}\frac{1}{{{x^2}\sqrt {1 + {x^2}} }} \cr
& {\text{Differentiating }\left[ { - \frac{{\sqrt {1 + {x^2}} }}{x} + C} \right] } \cr
& \underbrace {\frac{d}{{dx}}\left[ { - \frac{{\sqrt {1 + {x^2}} }}{x}} \right]}_{{\text{Use the quotient rule}}} + \frac{d}{{dx}}\left[ C \right] \cr
& {\text{computing derivatives}} \cr
& = - \frac{{x\left( {\frac{{2x}}{{2\sqrt {1 + {x^2}} }}} \right) - \sqrt {1 + {x^2}} \left( 1 \right)}}{{{x^2}}} + 0 \cr
& {\text{Simplifying}} \cr
& = - \frac{{\frac{{{x^2}}}{{\sqrt {1 + {x^2}} }} - \sqrt {1 + {x^2}} }}{{{x^2}}} \cr
& = - \frac{{{x^2} - 1 - {x^2}}}{{{x^2}\sqrt {1 + {x^2}} }} \cr
& = - \frac{{ - 1}}{{{x^2}\sqrt {1 + {x^2}} }} \cr
& =\frac{1}{{{x^2}\sqrt {1 + {x^2}} }} \cr
& {\text{The statement is true}}{\text{, then the formula is correct}}{\text{.}} \cr} $$
