Let $$\mathrm{f}(x)=\int \frac{\sqrt{x}}{(1+x)^2} \mathrm{~d} x, x \geq 0$$, then $$\mathrm{f}(3)-\mathrm{f}(1)$$ is equal to

Let $$\mathrm{f}(x)$$ be positive for all real $$x$$. If $$\mathrm{I}_1=\int_\limits{1-\mathrm{h}}^{\mathrm{h}} x \mathrm{f}(x(1-x)) \mathrm{d} x$$ and $$\mathrm{I}_2=\int_\limits{1-\mathrm{h}}^{\mathrm{h}} \mathrm{f}(x(1-x)) \mathrm{d} x$$, where $$(2 h-1)>0$$, then $$\frac{I_1}{I_2}$$ is

$$\int_\limits{-1}^3\left(\cot ^{-1}\left(\frac{x}{x^2+1}\right)+\cot ^{-1}\left(\frac{x^2+1}{x}\right)\right) \mathrm{d} x=$$

Let $$\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$$ and $$\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}$$ be continuous functions. Then the value of the integral $$\int_\limits{\frac{-\pi}{2}}^{\frac{\pi}{2}}[\mathrm{f}(x)+\mathrm{f}(-x)][\mathrm{g}(x)-\mathrm{g}(-x)] \mathrm{d} x$$ is