The value of $$\int\limits_{ - \pi }^\pi {{{{{\cos }^2}} \over {1 + {a^x}}}dx,\,\,a > 0,} $$ is

The value of integral, $$\int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx $$ is

The differential equation representing the family of curves $${y^2} = 2c\left( {x + \sqrt c } \right),$$ where $$c>0,$$ is a parameter, is of order and degree as follows:

If $$x{{dy} \over {dx}} = y\left( {\log y - \log x + 1} \right),$$ then the solution of the equation is :