If $a > 0, b > 0$ and $\int\dfrac{1}{ax^2 + b}dx = \dfrac{1}{\sqrt{6}}\tan^{-1}\left(\dfrac{\sqrt{2}x}{\sqrt{3}}\right) + c$, then $\int\dfrac{1}{bx^2 + a}dx = \ldots$

The value of integral $\int_0^1\cot^{-1}(1 + x^2 - x)\,dx$ is...

The value of $\int_{-\pi}^{\pi}\dfrac{2x(1 + \sin x)}{1 + \cos^2 x}dx$ is...

The area of the region enclosed by the lines $y = 2x, 2y = x$ and $x = 2$ (in sq.units) is...