If $\int \left\{ \cos^{-1} x - (1-x^2)^{-\frac{1}{2}} \right\} k \, dx = k \cdot \cos^{-1} x + c$,then $k = $ . . . . . . .

$\int \frac{1+\sin (\log x)}{1+\cos (\log x)} d x=$

$\int {{e^{{x^2}}}} \cdot {e^x}\left( {2{x^2} + x + 1} \right)dx = {e^{{x^2} + x}}\left( {f\left( x \right)} \right) + c$ where $c$ is the constant of integration. If the minimum value of $f(x)$ is equal to $m$,then find the value of $\left[ { - \frac{1}{m}} \right]$,where $[\cdot]$ denotes the Greatest Integer Function $(GIF)$.

The value of $\int e^x \left( \frac{x^2+4x+4}{(x+4)^2} \right) dx$ is

$\int 2^x (f^{\prime}(x) + f(x) \log 2) \, dx$ is equal to:

$\int_0^1 \frac{e^x(x - 1)}{(x + 1)^3} \, dx = $