Official solution
We can write,
7 - 6x - x2 = 16 - (x + 3)2
and \[{d \over {dx}}\left( {7 - 6x - {x^2}} \right) = - (2x + 6)\]
So, \[\int {{{2x + 5} \over {\sqrt {7 - 6x - {x^2}} }}} dx\]
= \[\int {{{2x + 6 - 1} \over {\sqrt {7 - 6x - {x^2}} }}} dx\]
= \[\int {{{2x + 6} \over {\sqrt {7 - 6x - {x^2}} }}} dx\] -
\[\int {{1 \over {\sqrt {16 - {{(x + 3)}^2}} }}} dx\]
= \[\int {{{2x + 6} \over {\sqrt {7 - 6x - {x^2}} }}} dx\] -
\[\int {{1 \over {\sqrt {{{\left( 4 \right)}^2} - {{(x + 3)}^2}} }}} dx\]
= \[ - 2\sqrt {7 - 6x - {x^2}} \] - \[{\sin ^{ - 1}}\left( {{{x + 3} \over 4}} \right)\] + C
By comparing with the given equation in the question we get,
A = - 2 and B = - 1