If   $${I_1} = \int_0^1 {{e^{ - x}}} {\cos ^2}x{\mkern 1mu} dx;$$

   $${I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x{\mkern 1mu} dx$$  and

$${I_3} = \int_0^1 {{e^{ - {x^3}}}} dx;$$ then

If    $$\int {{{2x + 5} \over {\sqrt {7 - 6x - {x^2}} }}} \,\,dx = A\sqrt {7 - 6x - {x^2}} + B{\sin ^{ - 1}}\left( {{{x + 3} \over 4}} \right) + C$$
(where C is a constant of integration), then the ordered pair (A, B) is equal to :

The curve satifying the differeial equation, (x² $$-$$ y²) dx + 2xydy = 0 and passing through the point (1, 1) is :

The sides of a rhombus ABCD are parallel to the lines, x $$-$$ y + 2 = 0 and 7x $$-$$ y + 3 = 0. If the diagonals of the rhombus intersect P(1, 2) and the vertex A (different from the origin) is on the y-axis, then the coordinate of A is :