Thomas' Calculus 13th Edition

$\frac{1}{3}(1-\frac{1}{x^2})^{3/2}+C$
We use the u-substitution: $u=1-\frac{1}{x^2}$ Derive: $du=\frac{2}{x^3}dx$ $\frac{1}{2}du=\frac{1}{x^3}dx$ Substitute: $\int \frac{1}{x^3}\sqrt{\frac{x^2 -1}{x^2}}dx$ =$\int \frac{1}{x^3}\sqrt{1-\frac{1}{x^2}}dx$ =$\int \sqrt{u}\frac{1}{2}du$ =$\frac{1}{2}\int u^{1/2}du$ =$\frac{1}{3}u^{3/2}+C$ =$\frac{1}{3}(1-\frac{1}{x^2})^{3/2}+C$