The equation of tangent to the curves $x = 1 - 3t^2$ and $y = t - 3t^3$ at the point $(-2, 2)$ is...

If $f(x) = \cos x$, then the value of $\int\dfrac{e^{f(x)}(x\sin^3 x + f(x))}{1 - (f(x))^2}dx = $

$\int\dfrac{x^4 + 1}{x^6 + 1}dx = $

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 $\int\dfrac{1}{bx^2 + a}dx = \ldots$