$\int e^{-x \log 2} 2^x dx =$

The velocity of a particle at time $t$ is given by the relation $v = 6t - \frac{t^2}{6}$. Its displacement $S$ is zero at $t = 0$,then the distance travelled in $3 \text{ s}$ is

If $f\left( \frac{3x - 4}{3x + 4} \right) = x + 2, x \ne -\frac{4}{3}$,and $\int f(x) dx = A \log |1 - x| + Bx + C$,then the ordered pair $(A, B)$ is equal to: (where $C$ is a constant of integration)

If $f'(x) = x^2 + 5$ and $f(0) = -1$,then $f(x) = $

If $f(\frac{x - 1}{x + 1}) = x + 1$, then $\int f(x) dx = \dots$

If $\int (\sin 2x + \cos 2x) dx = \frac{1}{\sqrt{2}} \sin (2x - c) + a$,then the value of $a$ and $c$ is: