The value of $$\int\limits_0^\infty {{{dx} \over {({x^2} + 4)({x^2} + 9)}}} $$ is

If $${I_1} = \int\limits_0^{\pi /4} {{{\sin }^2}xdx} $$ and $${I_2} = \int\limits_0^{\pi /4} {{{\cos }^2}xdx} $$, then

$$\int\limits_{ - 1}^4 {f(x)dx = 4} $$ and $$\int\limits_2^4 {\{ 3 - f(x)\} dx = 7} $$, then the value of $$\int\limits_{ - 1}^2 {f(x)dx} $$ is

$$\int\limits_0^{1000} {{e^{x - [x]}}dx} $$ is equal to