[m]dx=\frac{1}{cos^2t}dt[/m]
[m]1+x^2=1+tg^2t=\frac{1}{cos^2t}[/m]
[m] ∫ \frac{\sqrt{1+x^2}}{x^4}dx= ∫ \frac{\frac{1}{cos^2t}}{tg^4t}\cdot \frac{1}{cos^2t}dt= ∫ \frac{1}{sin^4t}dt=[/m]
[m] =∫ \frac{1}{sin^2t}\cdot \frac{1}{sin^2t} dt= ∫(1+ctg^2t)\cdot \frac{1}{sin^2t} dt= ∫\frac{1}{sin^2t} dt+ ∫ ctg^2t\cdot \frac{1}{sin^2t} dt=-ctgt-\frac{ctg^3t}{3}+C [/m]
Обратный переход
[m]x=tgt[/m] ⇒ [m]t=arctgx[/m]
[m]ctgt=\frac{1}{tgt}=\frac{1}{x}[/m]
[m]=-\frac{1}{x}-\frac{1}{3x^3}+C [/m]