Answer
$$2\sqrt {1 - x} + \ln \left| {\frac{{\sqrt {1 - x} - 1}}{{\sqrt {1 - x} + 1}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {1 - x} }}{x}} dx \cr
& {\text{Integrate by the substitution method}} \cr
& {\text{Let }}{u^2} = 1 - x,\,\,\,\,\,\,2udu = - dx \cr
& {\text{write in terms of }}u \cr
& \int {\frac{{\sqrt {1 - x} }}{x}} dx = \int {\frac{u}{{1 - {u^2}}}} \left( { - 2udu} \right) \cr
& = \int {\frac{{2{u^2}}}{{{u^2} - 1}}} du \cr
& = \int {\left( {2 + \frac{2}{{{u^2} - 1}}} \right)} du \cr
& \cr
& {\text{Decompose }}\frac{2}{{{u^2} - 1}}{\text{ into partial fractions}} \cr
& \frac{2}{{\left( {u + 1} \right)\left( {u - 1} \right)}} = \frac{A}{{u + 1}} + \frac{B}{{u - 1}} \cr
& {\text{Multiply by }}\left( {u + 1} \right)\left( {u - 1} \right){\text{ and simplify}} \cr
& 2 = A\left( {u - 1} \right) + B\left( {u + 1} \right) \cr
& {\text{if }}u = - 1,\,\,\,A = - 1 \cr
& {\text{if }}u = 1,\,\,\,B = 1 \cr
& \cr
& {\text{Replace the coefficients}} \cr
& \frac{2}{{\left( {u + 1} \right)\left( {u - 1} \right)}} = - \frac{1}{{u + 1}} + \frac{1}{{u - 1}} \cr
& \int {\left( {2 + \frac{2}{{{u^2} - 1}}} \right)} du = \int {\left( {2 - \frac{1}{{u + 1}} + \frac{1}{{u - 1}}} \right)} \cr
& \cr
& {\text{Integrate}} \cr
& = 2u - \ln \left| {u + 1} \right| + \ln \left| {u - 1} \right| + C \cr
& = 2u + \ln \left| {\frac{{u - 1}}{{u + 1}}} \right| + C \cr
& \cr
& {\text{write in terms of }}x \cr
& = 2\sqrt {1 - x} + \ln \left| {\frac{{\sqrt {1 - x} - 1}}{{\sqrt {1 - x} + 1}}} \right| + C \cr} $$
