Задача №1546

\[ \int\frac{dx}{x^4\sqrt{x^2-3}} =\left[x=\frac{\sqrt{3}}{\cos{z}};\;dx=\frac{\sqrt{3}\sin{z}dz}{\cos^2{z}};\;x^2-3=\frac{3\sin^2{z}}{\cos^2{z}}.\right] =\frac{\sgn(\cos{z})}{9}\cdot\int\cos^3{z}dz=\\ =\frac{\sgn(\cos{z})}{9}\cdot\int\left(1-\sin^2{z}\right)d(\sin{z}) =\sgn(\cos{z})\cdot\frac{\sin{z}}{27}\cdot\left(3-\sin^2{z}\right)+C=\\ =\frac{\sgn{x}}{27}\cdot\sqrt{1-\frac{3}{x^2}}\cdot\left(2+\frac{3}{x^2}\right)+C =\frac{\left(2x^2+3\right)\cdot\sqrt{x^2-3}}{27x^3}+C. \]

\[ \int\frac{dx}{x^4\sqrt{x^2-3}} =-\int\frac{d\left(\frac{1}{x}\right)}{x^2\sqrt{x^2-3}} =\left[t=\frac{1}{x}\right] =-\sgn{t}\cdot\int\frac{t^3dt}{\sqrt{1-3t^2}}=\\ =-\frac{\sgn{t}}{18}\cdot\int\frac{1-3t^2-1}{\sqrt{1-3t^2}}d\left(1-3t^2\right) =-\frac{\sgn{t}}{18}\cdot\int\left(\left(1-3t^2\right)^{\frac{1}{2}}-\left(1-3t^2\right)^{-\frac{1}{2}}\right)d\left(1-3t^2\right)=\\ =\frac{\sgn{t}}{27}\cdot\left(1-3t^2\right)^{\frac{1}{2}}\cdot\left(3t^2+2\right)+C =\frac{\sgn{x}}{27}\cdot\frac{\sqrt{x^2-3}}{|x|}\cdot\frac{3+2x^2}{x^2}+C =\frac{\left(2x^2+3\right)\cdot\sqrt{x^2-3}}{27x^3}+C. \]