The solution of the differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1+y^2}{1+x^2}$$ is

If $$\bar{a}, \bar{b}, \bar{c}$$ are three vectors such that $$\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}}+\overline{\mathrm{c}})+\overline{\mathrm{b}} \cdot(\overline{\mathrm{c}}+\overline{\mathrm{a}})+\overline{\mathrm{c}} \cdot(\overline{\mathrm{a}}+\overline{\mathrm{b}})=0 \quad$$ and $$\quad|\overline{\mathrm{a}}|=1$$, $$|\bar{b}|=8$$ and $$|\bar{c}|=4$$, then $$|\bar{a}+\bar{b}+\bar{c}|$$ has the value _________.

Let $$\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}$$ and $$\bar{b}=\hat{i}+\hat{j}$$. If $$\bar{c}$$ is a vector such that $$\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=|\overline{\mathrm{c}}|,|\overline{\mathrm{c}}-\overline{\mathrm{a}}|=2 \sqrt{2}$$ and the angle between $$(\bar{a} \times \bar{b})$$ and $$\bar{c}$$ is $$\frac{\pi}{6}$$, then $$|(\bar{a} \times \bar{b}) \times \bar{c}|$$ is

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