I Was Very Amazed At The Solution 5

Relevant wiki: Differentiation Under the Integral Sign

Let's generalize:

Let $\displaystyle F(\alpha) = \int \limits_{0}^1 \dfrac{x^{\alpha} - 1}{\ln x} \ dx$

Partial Differentiating both sides with respect to $\alpha$

$\displaystyle \begin{aligned} \implies F'(\alpha) & = \int \limits_0^1 \dfrac{\delta}{\delta \alpha } \left( \dfrac{x^{\alpha} - 1}{\ln x}\right) \ dx \\ F'(\alpha) & = \int \limits_0^1 \dfrac{1}{\ln x} x^{\alpha} \ln x \ dx = \int \limits_0^1 x^{\alpha} \ dx = \left. \dfrac{x^{\alpha + 1}}{\alpha + 1} \right|_0^1 \\ F'(\alpha) & = \dfrac{1}{\alpha + 1} \end{aligned}$

Integrating both sides with respect to $\alpha$ $\displaystyle \implies F(\alpha) = \ln (\alpha + 1) + C$

But as you notice, $\displaystyle F(0) = \int \limits_{0}^1 \dfrac{x^{0} - 1}{\ln x} \ dx = 0 = \ln (0 + 1) + C = C$

$\displaystyle \therefore F(\alpha) = \ln (\alpha + 1)$

In this problem, $\alpha = 2016$ $\displaystyle \implies F(2016) = \int \limits_{0}^1 \dfrac{x^{2016} - 1}{\ln x} \ dx = \ln (2016 + 1) = \ln (\mathbb{S})$

$\displaystyle \therefore \mathbb{S} = 2017 \implies \dfrac{\mathbb{S} + 1}{2} = \dfrac{2017 + 1}{2} = \boxed{1009}$