$\int {\frac{{\sqrt {1 + {x^2}} }}{{{x^4}}}} dx$

Let $I = \int {\frac{{\sqrt {1 + {x^2}} }}{{{x^4}}}} dx = \int {\frac{{\sqrt {1 + {x^2}} }}{x}} \cdot \frac{1}{{{x^3}}}dx$

$= \int {\sqrt {\frac{{1 + {x^2}}}{{{x^2}}}} } \cdot \frac{1}{{{x^3}}}dx = \int {\sqrt {\frac{1}{{{x^2}}} + 1} } \cdot \frac{1}{{{x^3}}}dx$

Let's put $1 + \frac{1}{{{x^2}}} = {t^2} \Rightarrow \frac{{ - 2}}{{{x^3}}}dx = 2tdt$

$\Rightarrow$ $- \frac{1}{{{x^3}}} = tdt$

therefore,$I = - \int {{t^2}} dt = - \frac{{{t^3}}}{3} + C = - \frac{1}{3}{\left( {1 + \frac{1}{{{x^2}}}} \right)^{3/2}} + C$