Partial to Last Year's Integrals

$\begin{aligned} I & = \int \frac {dx}{x^n-x} \\ & = \int \frac {dx}{x(x^{n-1}-1)} \\ & = \int \left(\frac {x^{n-2}}{x^{n-1}-1} - \frac 1x \right) dx \\ & = \frac {\ln(x^{n-1}-1)}{n-1} - \ln x + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \end{aligned}$

Therefore,

$\begin{aligned} \int_e^\infty \frac {dx}{x^{2017}-x} & = \frac {\ln(x^{2016}-1)}{2016} - \ln x\ \bigg|_e^\infty \\ & = \lim_{x \to \infty} \ln \left(\frac {x^{2016}-1}{x^{2016}}\right)^{\frac 1{2016}} - \frac {\ln(e^{2016}-1)}{2016} + \ln e \\ & = \lim_{x \to \infty} \ln \left(1 - \frac 1{x^{2016}}\right)^{\frac 1{2016}} - \frac {\ln(e^{2016}-1)}{2016} + 1 \\ & = 0 + 1 - \frac {\ln(e^{2016}-1)}{2016} \end{aligned}$

$\implies a+b+c+d = 1+2016+2016+1 = \boxed{4034}$ .