Answer
$\frac{3}{16}(2x-1)^{4/3}+\frac{3}{4}(2x-1)^{1/3}+C$
Work Step by Step
Let $u=2x-1=>x=\frac{1}{2}(u+1)=>dx=\frac{1}{2}du$
thus
=$\int \frac{x}{(2x-1)^{2/3}}$
=$\frac{\frac{1}{2}(u+1)}{u^{2/3}}(\frac{1}{2}du)$
=$\frac{1}{4}\int (u^{1/3}+u^{-2/3})du$
=$\frac{1}{4}(\frac{3}{4}u^{4/3}+3u^{1/3})+C$
=$\frac{3}{16}(2x-1)^{4/3}+\frac{3}{4}(2x-1)^{1/3}+C$
