Flip Side Of Feynman's Favourite Trick.

Relevant wiki: Gamma function & Factorial

$\begin{aligned} I & = \int_0^1 x^{2018} \ln^2 x \ dx & \small \color{#3D99F6} \text{Let }u = \ln x \implies e^u = x \implies e^u du = dx \\ & = \int_{-\infty}^0 e^{2019u} u^2 du & \small \color{#3D99F6} \text{Let }t = -2019u \implies dt = -2019\ du \\ & = \int_0^\infty \frac {t^2 e^{-t}}{2019^3} dt & \small \color{#3D99F6} \text{Gamma function }\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}\ dt \\ & = \frac {\Gamma (3)}{2019^3} & \small \color{#3D99F6} \text{Note that }\Gamma(n) = (n-1)! \\ & = \frac 2{2019^3} \end{aligned}$

Therefore, $a+b = 2+2019 = \boxed{2021}$ .

Feynman's trick makes use of the Liebniz Rule of Integration , more often than not the specific case $\frac{d}{dt} \int_{a}^{b} f(x,t)\,dx = \int_{a}^{b}\frac{\partial}{\partial t} f(x,t)\,dx$ where $a$ and $b$ are $\textbf{not}$ functions of $t$ . So the technique makes use of defining useful multi-variable integrals. First we must notice that $\frac{\partial}{\partial t} x^t = \ln(x)x^t, \, \, \, \, \, \frac{\partial}{\partial t} \ln(x)x^t = \ln^2(x)x^t.$ So if we let $I$ denote the integral in question, and
$f(t) = \int_0^1 x^t\,dx$ then the second derivative of $f$ evaluated at $t =2018$ will give us the value of the original integral $I$ : $I = f''(2018) = \frac{\partial^2}{\partial t^2} \Big|_{t=2018} \int_0^1 x^t \,dx = \frac{2}{(t+1)^3}\Big|_{t=2018} = \frac{2}{2019^3}$ $\frac{a}{b^3} = \frac{2}{2019^3} \, \,\therefore \, \, \boxed{a+b = 2021}.$ The twist is that usually one finds an integral that is easier to re-integrate back into an equivalent form of the original integral as a function of $t$ , but in this case we find that an integral partial differentiated twice is equivalent to the original integral.

Let $I(t)=\int_0^1 x^t dx= \frac {1}{(t+1)}$

Hence $I''(t) =\int_0^1 x^t (\ln x)^2 dx= \frac {2}{(t+1)^3}$

Put $t=2018$ to get $a=2$ and $b=2019$