Answer
$$\int\frac{1}{x^3}\sqrt{\frac{x^2-1}{x^2}}dx=\frac{1}{3}\Big(1-\frac{1}{x^2}\Big)^{3/2}+C$$
Work Step by Step
$$A=\int\frac{1}{x^3}\sqrt{\frac{x^2-1}{x^2}}dx$$ $$A=\int\frac{1}{x^3}\sqrt{1-\frac{1}{x^2}}dx$$
We set $u=1-\frac{1}{x^2}$
Then $$du=\Big(0-\frac{(1)'x^2-1(x^2)'}{x^4}\Big)dx=-\Big(\frac{0-2x}{x^4}\Big)dx$$ $$du=-\Big(-\frac{2}{x^3}\Big)dx=\frac{2}{x^3}dx$$
That means $$\frac{1}{x^3}dx=\frac{1}{2}du$$
Therefore, $$A=\frac{1}{2}\int\sqrt udu=\frac{1}{2}\int u^{1/2}du$$ $$A=\frac{1}{2}\times\frac{u^{3/2}}{\frac{3}{2}}+C=\frac{u^{3/2}}{3}+C$$ $$A=\frac{1}{3}\Big(1-\frac{1}{x^2}\Big)^{3/2}+C$$
