$\int e^x \left( \frac{x-1}{x^2} \right) dx =$

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

$\int \frac{e^{\cot x}}{\sin^2 x} (2 \log \csc x + \sin 2 x) dx =$

$\int {{e^x}\left[ {{{\sin }^{ - 1}}\frac{x}{a} + \frac{1}{{\sqrt {{a^2} - {x^2}} }}} \right]dx = }$

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 $4l+m=$

If $\int {{e^{{x^2}}}\left( {2 - \frac{1}{{{x^2}}}} \right)dx = {e^{{x^2}}}f(x) + C} $ and $f\left( {\frac{1}{2}} \right) = 2$,then $f(1)$ is equal to (where $C$ is an arbitrary constant).