## Thomas' Calculus 13th Edition

$${\sin ^{ - 1}}\left( {\frac{{x - 2}}{{\sqrt 3 }}} \right) + C$$
\eqalign{ & \int {\frac{{dx}}{{\sqrt { - {x^2} + 4x - 1} }}} \cr & {\text{complete the square for }} - {x^2} + 4x - 1 \cr & - {x^2} + 4x - 1 = - \left( {{x^2} - 4x + 4} \right) - 1 + 4 \cr & - {x^2} + 4x - 1 = 3 - {\left( {x - 2} \right)^2} \cr & {\text{then}}{\text{,}} \cr & = \int {\frac{{dx}}{{\sqrt {3 - {{\left( {x - 2} \right)}^2}} }}} \cr & {\text{use the substitution method}} \cr & u = x - 2,\,\,\,\,\,\,du = dx \cr & \int {\frac{{dx}}{{\sqrt {3 - {{\left( {x - 2} \right)}^2}} }}} = \int {\frac{{du}}{{\sqrt {{{\left( {\sqrt 3 } \right)}^2} - {u^2}} }}} \cr & {\text{integrating by the formula }}\cr &\int {\frac{1}{{\sqrt {{a^2} - {u^2}} }}du} = {\sin ^{ - 1}}\left( {\frac{u}{a}} \right) + C\cr &{\text{ with }}a = \sqrt 3 \cr & = {\sin ^{ - 1}}\left( {\frac{u}{{\sqrt 3 }}} \right) + C \cr & {\text{write in terms of }}x{\text{; replace }}u = x - 2 \cr & = {\sin ^{ - 1}}\left( {\frac{{x - 2}}{{\sqrt 3 }}} \right) + C \cr}