$\int {{e^x}(1 + \tan x + {{\tan }^2}x)\,dx = } $

Evaluate the integral: $\int {\frac{{{e^{{{\tan }^{ - 1}}x}}}}{{(1 + {x^2})}}\,\,\left[ {{{\left( {{{\sec }^{ - 1}}\,\sqrt {1 + {x^2}} } \right)}^2}\,\, + \,\,{{\cos }^{ - 1}}\,\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)} \right]} \,\,\,dx$ for $x > 0$.

The value of $\int \frac{(x-1) e^x}{(x+1)^3} \,d x$ is equal to

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

Integrate the function: $\frac{x e^{x}}{(1+x)^{2}}$

If $\int \frac{3-x^2}{1-2 x+x^2} e^x d x=e^x f(x)+c$,then $f(x)$ is: