Who's up to the challenge? 2

Let $I$ be the integral, then:

$\begin{aligned} I & = \int_1^2 \frac{dx}{\color{#3D99F6}{5x^2+2x+5}} \quad \quad \quad \quad \small \color{#3D99F6}{\text{See Note.}} \\ & = \int_1^2 \frac{dx}{\color{#3D99F6} {\frac{24}{5} \left[1 + \left(\frac{5x+1}{\sqrt{24}} \right)^2 \right]}} \quad \quad \small \color{#3D99F6}{\text{Let } \tan \theta = \frac{5x+1}{\sqrt{24}} \Rightarrow \sec^2 \theta \space d\theta = \frac{5}{\sqrt{24}} dx} \\ & = \frac{5}{24} \dot{} \frac{\sqrt{24}}{5} \int_{\tan^{-1} \frac{6}{\sqrt{24}}} ^{\tan^{-1} \frac{11}{\sqrt{24}}} \frac{\sec^2 \theta}{\sec^2 \theta} d\theta \\ & = \frac{1}{\sqrt{24}} \left[ \theta \right]_{\tan^{-1} \frac{6}{\sqrt{24}}} ^{\tan^{-1} \frac{11}{\sqrt{24}}} \\ & = \frac{1}{\sqrt{24}} \left( \tan^{-1} \frac{11}{\sqrt{24}} - \tan^{-1} \frac{6}{\sqrt{24}} \right) \\ & = \frac{1}{2\sqrt{6}} \tan^{-1} \left(\frac{\frac{5}{\sqrt{24}}}{1+ \frac{66}{24}} \right) \\ & = \frac{1}{2\sqrt{6}} \tan^{-1} \left(\frac{5 \sqrt{24}}{90} \right) \\ & = \frac{1}{2\sqrt{6}} \cot^{-1} \left(\frac{18}{\sqrt{24}} \right) \\ & = \frac{1}{2\sqrt{6}} \cot^{-1} \left(3 \sqrt{\frac{3}{2}} \right) \end{aligned}$

$\Rightarrow a+b+c+d = 3+3+2+6 = \boxed{14}$

$$

$\color{#3D99F6}{\text{Note:}}$

$\begin{aligned} 5x^2+2x+5 & = 5\left(x^2 + \frac{2x}{5} + \frac{1}{25} \right) - \frac{1}{5} + 5 \\ & = 5\left(x + \frac{1}{5} \right)^2 + \frac{24}{5} \\ & = \frac{\left(5x + 1 \right)^2}{5} + \frac{24}{5} \\ & = \frac{24}{5} \left(\frac{(5x+1)^2}{24} +1 \right) \end{aligned}$