The solution of $$(1+x y) y d x+(1-x y) x d y=0$$ is

The integral $$\int_\limits{\pi / 6}^{\pi / 3} \sec ^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x d x$$ is equal to

$$\lim _\limits{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 \frac{x}{2}}=$$

$$A(1,-3), B(4,3)$$ are two points on the curve $$y=x-\frac{4}{x}$$. The points on the curve, the tangents at which are parallel to the chord $$A B$$, are