The value of $ \int \frac{e^{x}\left(x^{2} \tan ^{-1} x+\tan ^{-1} x+1\right)}{x^{2}+1} d x $ is equal to

$ \int e^{\sin x} \cdot \left(\frac{\sin x+1}{\sec x}\right) d x $ is equal to

$\int e^x \left( \frac{x+5}{(x+6)^2} \right) dx$ is equal to:

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

Evaluate the integral: $\int \frac{\log x}{(1+\log x)^2} dx$

$\int \frac{\log x}{(1 + \log x)^2} dx = $