## Thomas' Calculus 13th Edition

$$- \sqrt {1 - {x^2}} + {\sin ^{ - 1}}x + C$$
\eqalign{ & \int {\sqrt {\frac{{x + 1}}{{1 - x}}} } dx \cr & {\text{Use the radical properties and rationalize the numerator}} \cr & = \int {\frac{{\sqrt {x + 1} }}{{\sqrt {1 - x} }}} dx \cr & = \int {\frac{{\sqrt {x + 1} }}{{\sqrt {1 - x} }}} \left( {\frac{{\sqrt {x + 1} }}{{\sqrt {x + 1} }}} \right)dx = \int {\frac{{{{\left( {\sqrt {x + 1} } \right)}^2}}}{{\sqrt {\left( {1 - x} \right)\left( {1 + x} \right)} }}} dx \cr & = \int {\frac{{x + 1}}{{\sqrt {1 - {x^2}} }}} dx \cr & {\text{Using Trigonometric Substitutions}} \cr & {\text{We set}}{\text{, }}x = \sin \theta,{\text{ }}dx = \cos \theta d\theta \cr & {\text{With these substitutions}}{\text{, we have}} \cr & \int {\frac{{x + 1}}{{\sqrt {1 - {x^2}} }}} dx = \int {\frac{{\sin \theta + 1}}{{\sqrt {1 - {{\sin }^2}\theta } }}} \left( {\cos \theta } \right)d\theta \cr & {\text{use the fundamental identity }}{\cos ^2}\theta + {\sin ^2}\theta = 1 \cr & = \int {\frac{{\sin \theta + 1}}{{\sqrt {{{\cos }^2}\theta } }}} \left( {\cos \theta } \right)d\theta \cr & = \int {\left( {\sin \theta + 1} \right)} d\theta \cr & {\text{integrate}} \cr & = - \cos \theta + \theta + C \cr & {\text{write in terms of }}x,{\text{ }}\cot \theta = \sqrt {1 - {x^2}} {\text{ and }}\theta = {\sin ^{ - 1}}x \cr & = - \sqrt {1 - {x^2}} + {\sin ^{ - 1}}x + C \cr}