Answer
$$ - 1 + \frac{\pi }{2}$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^0 {\sqrt {\frac{{1 + y}}{{1 - y}}} } dy \cr
& {\text{Use the property }}\sqrt {\frac{a}{b}} = \frac{{\sqrt a }}{{\sqrt b }} \cr
& \int_{ - 1}^0 {\frac{{\sqrt {1 + y} }}{{\sqrt {1 - y} }}} dy \cr
& {\text{Rationalize the numerator}} \cr
& = \int_{ - 1}^0 {\frac{{\sqrt {1 + y} }}{{\sqrt {1 - y} }}} \left( {\frac{{\sqrt {1 + y} }}{{\sqrt {1 + y} }}} \right)dy \cr
& = \int_{ - 1}^0 {\frac{{1 + y}}{{\sqrt {1 - {y^2}} }}} dy \cr
& = \int_{ - 1}^0 {\left( {\frac{1}{{\sqrt {1 - {y^2}} }} + \frac{y}{{\sqrt {1 - {y^2}} }}} \right)} dy \cr
& {\text{Integrating}} \cr
& \left( {{{\sin }^{ - 1}}y - \sqrt {1 - {y^2}} } \right)_{ - 1}^0 \cr
& {\text{Evaluating, we get:}} \cr
& = \left( {{{\sin }^{ - 1}}0 - \sqrt {1 - {0^2}} } \right) - \left( {{{\sin }^{ - 1}}\left( { - 1} \right) - \sqrt {1 - {{\left( { - 1} \right)}^2}} } \right) \cr
& = \left( {0 - 1} \right) - \left( { - \frac{\pi }{2} - \sqrt 0 } \right) \cr
& = - 1 + \frac{\pi }{2} \cr} $$
