If $\int e^x \left( \frac{x^2-8x+19}{(x-1)^5} \right) dx = \frac{e^x(lx+m)}{(x-1)^4} + C$, then $4l+m=$

$\int {{e^{{x^2}}}} \cdot {e^x}\left( {2{x^2} + x + 1} \right)dx = {e^{{x^2} + x}}\left( {f\left( x \right)} \right) + c$ 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 $(GIF)$.

Find : $\int \frac{(x^{2}+1) e^{x}}{(x+1)^{2}} d x$

$\int \frac{(\log x-1)^2}{\left[1+(\log x)^2\right]^2} d x=$ (where $C$ is constant of integration.)

$\int e^{x \operatorname{cosec} x} \cdot \operatorname{cosec} x \cdot(1-x \cot x) \, dx =$

$\int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx = $ . . . . . . $+ c$.