While this is

Or this could be done like this !! ;-) $\mathfrak{I}=\displaystyle \int _{ 3 }^{ 37} \frac { 1+x+{ x }^{ 2 } }{ 1+{ x }^{ 2 } } dx$ $=\displaystyle \int _{ 3 }^{ 37 }( 1+ \frac { x}{1+x^2})dx$ $=34+\displaystyle \int _{ 3 }^{ 37 } \frac{1}{2}(\frac { d(1+x^2)}{1+x^2})$ $=34+\dfrac{1}{2}(\ln (\frac{1370}{10}))\approx 36.46$ Hence $\left\lfloor A \right\rfloor =\boxed{\large 36}$

$\begin{aligned} A & = \int_3^{37} \frac{1+x+x^2}{1+x^2} \space dx \quad \quad \small \color{#3D99F6}{\text{Let } x = \tan \theta \quad \Rightarrow dx = \sec^2 \theta \space d\theta} \\ & = \int_{\tan^{-1} 3}^{\tan^{-1} 37} \frac{\left(1+\tan \theta + \tan^2 \theta \right) \sec^2 \theta}{\sec^2 \theta} \space d\theta \\ & = \int_{\tan^{-1} 3}^{\tan^{-1} 37} \left(\tan \theta + \sec^2 \theta \right) \space d\theta \\ & = \int_{\tan^{-1} 3}^{\tan^{-1} 37} \frac{\sin \theta}{\cos \theta} \space d\theta + \int_{\tan^{-1} 3}^{\tan^{-1} 37} \sec^2 \theta \space d\theta \\ & = - \int_{\frac{1}{\sqrt{10}}}^{\frac{1}{\sqrt{1370}}} \frac{d \cos \theta}{\cos \theta} + \int_3^{37} d \tan \theta \space d\theta \\ & = \left[\ln (\cos \theta) \right]^{\frac{1}{\sqrt{10}}}_{\frac{1}{\sqrt{1370}}} + \left[ \tan \theta \right]_3^{37} \\ & = \frac{\ln 137}{2} + 34 \approx 36.460 \end{aligned}$

$\Rightarrow \lfloor A \rfloor = \boxed{36}$