Answer
$\frac{2}{27}(1-\frac{3}{x^3})^{3/2}+C$
Work Step by Step
Let u=$1-\frac{3}{x^3}=>du=\frac{9}{x^4}dx=>\frac{1}{9}du=\frac{1}{x^4}dx$
=$\int \sqrt{\frac{x^3-3}{x^{11}}}dx$
=$\int \frac{1}{x^4} \sqrt{\frac{x^3-3}{x^{3}}}dx$
=$\int \frac{1}{x^4} \sqrt{1-\frac{3}{x^3}}dx$
=$\int \sqrt{u}\frac{1}{9}du$
=$\frac{1}{9} \int u^{1/2}du$
=$\frac{2}{27}u^{3/2}+C$
=$\frac{2}{27}(1-\frac{3}{x^3})^{3/2}+C$
