For $$x>1$$, if $$(2 x)^{2 y}=4 \mathrm{e}^{2 x-2 y}$$, then $$(1+\log 2 x)^2 \frac{\mathrm{d} y}{\mathrm{~d} x}$$ is equal to

The vectors are $$\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \bar{b}=\hat{i}+\hat{j}$$. If $$\bar{c}$$ is a vector such that $$\bar{a} \cdot \bar{c}=|\bar{c}|$$ and $$|\bar{c}-\bar{a}|=2 \sqrt{2}$$, angle between $$\bar{a} \times \bar{b}$$ and $$\bar{c}$$ is $$\frac{\pi}{4}$$, then $$|(\bar{a} \times \bar{b}) \times \bar{c}|$$ is

$$\int \frac{1}{7-6 x-x^2} d x=$$

$$\int \frac{d x}{\sin x+\cos x}=$$