The function $f(x) = x(x + 3)e^{-\left(\frac{1}{2}\right)x}$ satisfies all the conditions of Rolle's theorem in $[-3, 0]$, then $c =$

If $\int f(x)dx = g(x) + c$ then $\int f^{-1}(x)dx =$

$\int\dfrac{\log x}{(1 + \log x)^2}dx =$

$\int\dfrac{1}{\sqrt{2x - x^2}}dx =$