$\int \left( \frac{x^2+1}{(x+1)^2} \right) e^x \, dx = \text{ . . . . . . }$.

$\int {{e^{{x^2}}}} \cdot {e^x}\left( {2{x^2} + x + 1} \right)dx = {e^{{x^2} + x}}\left( {f\left( x \right)} \right) + c$ 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 $\int e^{x} \left[ \frac{1+\sin x}{1+\cos x} \right] dx$ is

Let $f(t) = \int \left( \frac{1 - \sin(\ln t)}{1 - \cos(\ln t)} \right) dt$,for $t > 1$. If $f(e^{\pi/2}) = -e^{\pi/2}$ and $f(e^{\pi/4}) = \alpha e^{\pi/4}$,then $\alpha$ equals:

$\int_{\alpha+1}^{\alpha} \frac{e^x(\alpha-x)}{(x-\alpha+1)^2} dx =$

$\int e^x \cdot \sec x(1+\tan x) \, dx = $ . . . . . . $+ C$.