Интеграл от sqrt((x- 4)/(x^3))

Решите задачу:

$\it\displaystyle \int\sqrt\frac{x-4}{x^3}dx=\int\frac{1}{x}\sqrt\frac{x-4}{x}dx=-2\int\frac{t^2}{t^2-1}dt=\\=-2\int\frac{t^2-1+1}{t^2-1}dt=-2\int(1+\frac{1}{t^2-1})dt=\\=-2t-ln|\frac{t-1}{t+1}|+C=-2\sqrt\frac{x-4}{x}-ln|\frac{\sqrt{x-4}-\sqrt x}{\sqrt{x-4}+\sqrt x}|+C=\\=-2\sqrt\frac{x-4}{x}-ln|\frac{\sqrt{x(x-4)}+2-x}{2}|=\\=-2\sqrt\frac{x-4}{x}-ln|\sqrt{x(x-4)}+2-x|+C^*$

$\it\displaystyle\frac{x-4}{x}=t^2\\x-4=xt^2\\-4=xt^2-x\\-4=x(t^2-1)\\x=-\frac{4}{t^2-1}\\\frac{1}{x}=-\frac{1}{4}(t^2-1)\\dx=\frac{8t}{(t^2-1)^2}dt$