$$ \int_1^2 x \sqrt{4-x^2} d x= $$

If $[x]$ denotes the greatest integer function of $x$ and

$$ \int_{-3 / 2}^{3 / 2}[2 x-3] d x=k, \text { then }\left|k+\frac{1}{2}\right|= $$

$$ \int_1^4\left(x+\sqrt{x}+\frac{1}{x}\right) d x-\int_1^{2 \log 2} d x= $$

Let $I=\int_{-\pi / 4}^{\pi / 4} \frac{1}{2-\cos 2 x}\left(\frac{\beta}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) d x$. Given that $\int \frac{d x}{1+k x^2}=\frac{1}{\sqrt{k}} \tan ^{-1}(\sqrt{k} x)+c, \tan ^{-1}(0)=0$ and $\tan ^{-1}(\sqrt{3})=\pi / 3$. Then, $3 I^2=$