Inte gration

Note that the integral does not rely on x, so we can take out the constant and get an integral value of 2018-2017=1. As for the constant we get as follows.

$\frac{18!+17!}{18!-17!} = \frac{18\cdot 17!+17!}{18\cdot 17!-17!} = \frac{17!(18+1)}{17!(18-1)}=\frac{18+1}{18-1}=\frac{19}{17}=\frac{a}{b}$

Thus our required answer is $19+17=36$

$\displaystyle \int_{2017}^{2018} \frac {18!+17!}{18!-17!}\ dx = \int_{2017}^{2018} \frac {17!(18+1)}{17!(18-1)}\ dx = \int_{2017}^{2018} \frac {19}{17}\ dx = \frac {19}{17}\ \bigg|_{2017}^{2018} = \frac {19}{17}$

Therefore, $a+b= 19+17 = \boxed{36}$ .