Answer
$$\ln \left| {\frac{{\sqrt x - 1}}{{\sqrt x + 1}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{x^{3/2}} - \sqrt x }}} dx \cr
& {\text{Factor the denominator}} \cr
& = \int {\frac{1}{{\sqrt x \left( {x - 1} \right)}}} dx \cr
& {\text{let }}t = \sqrt x ,\,\,\,\,dt = \frac{1}{{2\sqrt x }}dx,\,\,\,\,\,2tdt = dx \cr
& = \int {\frac{1}{{t\left( {{t^2} - 1} \right)}}} \left( {2tdt} \right) \cr
& = \int {\frac{2}{{{t^2} - 1}}} dt \cr
& {\text{Decompose the integrand into partial fractions}} \cr
& \frac{2}{{{t^2} - 1}} = \frac{2}{{\left( {t + 1} \right)\left( {t - 1} \right)}} \cr
& \frac{2}{{\left( {t + 1} \right)\left( {t - 1} \right)}} = \frac{A}{{t + 1}} + \frac{B}{{t - 1}} \cr
& {\text{Multiply by }}\left( {t + 1} \right)\left( {t - 1} \right){\text{ and simplify}} \cr
& 2 = A\left( {t - 1} \right) + B\left( {t + 1} \right) \cr
& {\text{if }}t = - 1,\,\,\,A = - 1 \cr
& {\text{if }}t = 1,\,\,\,B = 1 \cr
& \cr
& {\text{Replace the coefficients}} \cr
& \frac{2}{{\left( {t + 1} \right)\left( {t - 1} \right)}} = - \frac{1}{{t + 1}} + \frac{1}{{t - 1}} \cr
& \cr
& \int {\frac{2}{{{t^2} - 1}}dt} = \int {\left( { - \frac{1}{{t + 1}} + \frac{1}{{t - 1}}} \right)dt} \cr
& {\text{Integrate}} \cr
& = - \ln \left| {t + 1} \right| + \ln \left| {t - 1} \right| + C \cr
& = \ln \left| {\frac{{t - 1}}{{t + 1}}} \right| + C \cr
& {\text{Replace }}t = \sqrt x \cr
& = \ln \left| {\frac{{\sqrt x - 1}}{{\sqrt x + 1}}} \right| + C \cr} $$
