Answer
$$\int \frac{x}{(2x-1)^{2/3}}dx=\frac{3}{16}(2x-1)^{4/3}+\frac{3}{4}(2x-1)^{1/3}+C$$
Work Step by Step
$$A=\int \frac{x}{(2x-1)^{2/3}}dx$$
We set $u=2x-1$. That means $x=\frac{u+1}{2}$
Then $$du=2dx$$ $$dx=\frac{1}{2}du$$
Therefore, $$A=\frac{1}{2}\int\frac{\frac{u+1}{2}}{u^{2/3}}du=\frac{1}{2}\int \frac{u+1}{2u^{2/3}}du$$ $$A=\frac{1}{4}\int\frac{u+1}{u^{2/3}}du$$ $$A=\frac{1}{4}\int\Big(u^{1/3}+\frac{1}{u^{2/3}}\Big)du=\frac{1}{4}\int\Big(u^{1/3}+u^{-2/3}\Big)du$$ $$A=\frac{1}{4}\Big(\frac{u^{4/3}}{\frac{4}{3}}+\frac{u^{1/3}}{\frac{1}{3}}\Big)+C$$ $$A=\frac{3u^{4/3}}{16}+\frac{3}{4}u^{1/3}+C$$ $$A=\frac{3}{16}(2x-1)^{4/3}+\frac{3}{4}(2x-1)^{1/3}+C$$
