If $f(x) = \begin{cases} 1 - 2x, & x \leq 0 \\ 1 + 2x, & x > 0 \end{cases}$, then $\int_{-1}^{1} f(x) \, dx =$

If $\int_0^a 3x^2dx = 8$, then the value of $a$ is:

If $\displaystyle\int_0^{40} \dfrac{dx}{2x+1} = \log_e k$, then the value of $k$ is :

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{tanx}}$ is equal to

$\int \frac{x}{(x-1)(x-2)} dx =$

(where c is an arbitrary constant)

$\int \frac{\sqrt{\tan x}}{\sin x \cos x} dx$ equals :

If $\displaystyle\int_0^3 (8x^3+5x^2-2x-k)\,dx = 3$, where $k$ is constant, then $k$ is equal to