Answer
$$\int\sqrt{\frac{x-1}{x^5}}dx=\frac{2}{3}\Big(1-\frac{1}{x}\Big)^{3/2}+C$$
Work Step by Step
$$A=\int\sqrt{\frac{x-1}{x^5}}dx=\int\sqrt{\frac{1}{x^4}\times\frac{x-1}{x}}dx$$ $$A=\int\frac{1}{x^2}\sqrt{1-\frac{1}{x}}dx$$
We set $u=1-\frac{1}{x}$
Then $$du=0-\Big(-\frac{1}{x^2}\Big)dx=\frac{1}{x^2}dx$$
Therefore, $$A=\int\sqrt udu=\int u^{1/2}du$$ $$A=\frac{u^{3/2}}{\frac{3}{2}}+C=\frac{2u^{3/2}}{3}+C$$ $$A=\frac{2}{3}\Big(1-\frac{1}{x}\Big)^{3/2}+C$$