Integral ( I was too lazy to make a good name... )

$\displaystyle Let ~t=x+2.~\\ \displaystyle \therefore~ \int _{ 1 }^{ 2 }{ \frac { x }{ { x }^{ 2 }+4x+8 } } dx\\ \displaystyle = \int _{ 3 }^{ 4 }{ \frac { t-2 }{ { t }^{ 2 }+2^2 } } dt \\ \displaystyle = \int _{ 3 }^{ 4 }{ \frac { t }{ { t }^{ 2 }+2^2 } } dt - \int _{ 3 }^{ 4 }{ \frac { 2 }{ { t }^{ 2 }+2^2 } } dt\\ \displaystyle \int _{ 3 }^{ 4 }{ \frac { 2 }{ { t }^{ 2 }+2^2 } } dt\\ \displaystyle =2*\dfrac 1 2 *Tan^{-1}\frac 4 2 - 2*\dfrac 1 2 *Tan^{-1}\frac 3 2 \\ \displaystyle =Tan^{-1} 2 - Tan^{-1}\frac 3 2..........................(1)\\ \displaystyle Let~~ u=t^2. \implies~ \dfrac 1 2 du=t*dt~~and \\ \displaystyle = \int _{ 3 }^{ 4 }{ \frac { t }{ { t }^{ 2 }+2^2 } } dt \\\displaystyle =\dfrac 1 2 * \int _{ 9 }^{ 16 } \dfrac {1 }{ u+4 } du \\\displaystyle =\dfrac 1 2 *Ln \dfrac{20}{13} ...........(2) \\ \displaystyle \therefore~ \int _{ 1 }^{ 2 }{ \frac { x }{ { x }^{ 2 }+4x+8 } } dx \\ \displaystyle =(2)~-~(1)\\ \displaystyle =\dfrac 1 2 *Ln \dfrac{20}{13}~-~Tan^{-1} 2 - Tan^{-1}\frac 3 2 ~=~\arctan { \frac { a }{ b } } -\arctan { c } +\frac { 1 }{ d } \ln { \frac { e }{ f } }\\ \therefore~a+b+c+d+e+f=3+2+2+2+20+13=~~~\Huge \color{#D61F06}{42}$

Let $I$ be the integration. Then, $\begin{aligned} I & = \displaystyle\int_{1}^{2}\frac{x}{x^2+4x+8}\,dx \\& = \displaystyle\int_{1}^{2}\frac{x+2-2}{x^2+4x+8}\,dx\\& = \displaystyle\int_{1}^{2}\frac{x+2}{x^2+4x+8}\,dx -\displaystyle\int_{1}^{2}\frac{2}{x^2+4x+8}\,dx\\& = \frac{1}{2}\ln(x^2+4x+8)\bigg|_{1}^{2}-\displaystyle\int_{1}^{2}\frac{2}{(x+2)^2+2^2}\,dx\\& = \frac{1}{2}\ln\left(\frac{20}{13}\right) - \left[\tan ^{-1}\frac{x+2}{2}\right]_{1}^{2}\end{aligned}$ Setting the upper and lower limits we obtain $I =\tan ^{-1}\left(\frac{3}{2}\right)-\tan ^{-1}(2) +\frac{1}{2}\ln\left(\frac{20}{13}\right)$ Thus ,the sum of $\small a+b+c+d+e+f \implies 3+2+2+2+20+13\implies \boxed{42}$

@Seth Lovelace The best ever possible name to this problem - The Answer to Life, Universe and Everything ! .. sincerely hope that you get the reference.

nice solution