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

If $\int \dfrac{x+1}{x^2+1}dx = \tan^{-1}x + g(x) + c$, where $c$ is constant of integration, then the function $g(x)$ is monotonically increasing in the interval

$\int \sqrt{1+\sin x}\,dx$ is equal to

If $\int_0^a (x^2 - 4x + 1)dx = 6$, then the real value of $a$ is