If $$\int_\limits0^{\frac{1}{2}} \frac{x^2}{\left(1-x^2\right)^{\frac{3}{2}}} \mathrm{~d} x=\frac{\mathrm{k}}{6}$$, then the value of $$\mathrm{k}$$ is

$$\int_\limits1^2 \frac{\mathrm{d} x}{\left(x^2-2 x+4\right)^{\frac{3}{2}}}=\frac{\mathrm{k}}{\mathrm{k}+5} \text {, then } \mathrm{k} \text { has the value }$$

If $$\mathrm{f}(x)$$ is a function satisfying $$\mathrm{f}^{\prime}(x)=\mathrm{f}(x)$$ with $$\mathrm{f}(0)=1$$ and $$\mathrm{g}(x)$$ is a function that satisfies $$\mathrm{f}(x)+\mathrm{g}(x)=x^2$$. Then the value of the integral $$\int_\limits0^1 f(x) g(x) d x$$ is

$$\int_\limits{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{\mathrm{d} x}{1+\cos x}$$ is equal to