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

$\int_0^1 \frac{x e^x}{(x+1)^2} d x$ is equal to

If $\int e^x \left(\frac{x+2}{x+4}\right)^2 dx = f(x) + C$,then $f(x) =$

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

Integrate the function: $e^{x} \left( \frac{1+\sin x}{1+\cos x} \right)$

Evaluate the definite integral $\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) e^{2 x} d x$.