Answer
$$\int\frac{1}{x^{3/2}-\sqrt x}dx=\ln|\sqrt x-1|-\ln|\sqrt x+1|+C$$
Work Step by Step
$$I=\int\frac{1}{x^{3/2}-\sqrt x}dx=\int\frac{1}{\sqrt x(x-1)}dx$$
We set $u=\sqrt x$, so $$du=\frac{1}{2\sqrt x}dx$$ $$\frac{1}{\sqrt x}dx=2du$$
Also, $x-1=u^2-1$
$$I=\int\frac{2}{u^2-1}du=\int\frac{2}{(u-1)(u+1)}du$$ $$I=\int\frac{(u+1)-(u-1)}{(u-1)(u+1)}du$$ $$I=\Big(\int\frac{du}{u-1}-\int\frac{du}{u+1}\Big)$$ $$I=\ln|u-1|-\ln|u+1|+C$$ $$I=\ln|\sqrt x-1|-\ln|\sqrt x+1|+C$$
