$\int {{e^x}(1 + \tan x + {{\tan }^2}x)\,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 $\int e^x\left(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}+\frac{x}{1-x^2}\right) d x=g(x)+C$ where $C$ is the constant of integration,then $g \left(\frac{1}{2}\right)$ equals :

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

Evaluate the integral: $\int {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}} \left( {x + \sqrt x } \right)dx$

$\int e^x(1-\cot x+\cot^2 x) dx =$