Integrals — Class 12 Maths Solution

: Let $I = \int {\cfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx}$

$= \int {\cfrac{1}{{\sqrt {7 - \left( {{x^2} + 6x} \right)} }}dx} = \int {\cfrac{{dx}}{{\sqrt {16 - \left( {{x^2} + 6x + 9} \right)} }}}$

$= \int {\cfrac{1}{{\sqrt {{{\left( 4 \right)}^2} - {{\left( {x + 3} \right)}^2}} }}dx} = {\sin ^{ - 1}}\left( {\cfrac{{x + 3}}{4}} \right) + C$