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