Double Integral For Days

$\begin{aligned} I & = \int_0^1 \int_0^1 \left(\frac {(x-1)e^y}{\ln x}\sum_{n=0}^{2018} x^n \right)dx \ dy \\ & = \int_0^1 e^y \ dy \cdot \int_0^1 \left(\frac {x-1}{\ln x}\sum_{n=0}^{2018} x^n \right)dx \\ & = e^y \bigg|_0^1 \cdot \int_0^1 \frac {x-1}{\ln x}\cdot \frac {x^{2019}-1}{x-1} dx \\ & = (e-1) \color{#3D99F6} \int_0^1 \frac {x^{2019}-1}{\ln x} dx & \small \color{#3D99F6} \text{See note,} \\ & = (e-1) \color{#3D99F6} \ln (2020) \\ & \approx \boxed{13} \end{aligned}$

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Note:

$\begin{aligned} J(a) & = \int_0^1 \frac {x^a - 1}{\ln x} dx \\ \frac {\partial J(a)}{\partial a} & = \int_0^1 \frac {x^a\ln x}{\ln x} dx = \int_0^1 x^a \ dx = \frac {x^{a+1}}{a+1} \bigg|_0^1 = \frac 1{a+1} \\ \implies J(a) & = \int \frac 1{a+1} da = \ln (a+1) +\color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ J(0) & = \ln (0+1) + C = 0 & \small \color{#3D99F6} \implies C = 0 \\ \implies J(a) & = \ln(a+1) \\ J(2019) & = \ln (2020) \end{aligned}$