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

The area (in square units) in the first quadrant bounded by the curve $y=x^2+2$ and the lines $y=x+1, x=0$ and $x=3$, is

The co-ordinates of the point where the line through $\mathrm{A}(3,4,1)$ and $\mathrm{B}(5,1,6)$ crosses the $x y$-plane are

The value of $\frac{\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)}{\operatorname{cosec}^{-1}(-\sqrt{2})+\cos ^{-1}\left(\frac{-1}{2}\right)}$