Do you know the best way to integrate this?

Similar solution to @Noel Lo 's, perhaps better presentation.

$\begin{aligned} I & = \int_0^1 x^2 \tan^{-1} x \ dx \quad \quad \small \color{#3D99F6}{\text{By integration by parts, }f' = x^2, \ g = \tan^{-1}x} \\ & = \frac {x^3}3 \tan^{-1}x \bigg|_0^1 - \frac 13 \int_0^1 \frac {x^3}{1+x^2} dx \\ & = \frac 13 \frac \pi 4 - \frac 13 \int_0^1 \frac {x+x^3-x}{1+x^2} dx \\ & = \frac \pi{12} - \frac 13 \int_0^1 \left(x - \frac {x}{1+x^2} \right) dx \\ & = \frac \pi{12} - \frac 13 \left[ \frac {x^2}2 - \frac 12 \ln (1+x^2) \right] _0^1 \\ & = \frac \pi{12} - \frac {1-\ln 2}6 \end{aligned}$

$\implies A+B+C+D = 12+1+2+6 = \boxed{21}$

$\int_{0}^{1} x^2 \tan^{-1} x \,dx =[\int_{0}^{1} x^2 \,dx] [\tan^{-1} x]_{0}^{1} - \int_{0}^{1}[\int \, x^2 \,dx][\dfrac{d}{dx}\tan^{-1} x] \,dx$ $=[\dfrac{x^3}{3}]_{0}^{1} [\tan^{-1} x]_{0}^{1} - \int_{0}^{1}[\frac{x^3}{3}][\dfrac{1}{1+x^2}]\,dx = [\dfrac{1}{3}][\dfrac{\pi}{4}] - \dfrac{1}{3} \int_{0}^{1}[\dfrac{x^3}{1+x^2}]\,dx$ $=\dfrac{\pi}{12} - \dfrac{1}{3} \int_{0}^{1}[x-\dfrac{x}{1+x^2}]\,dx=\dfrac{\pi}{12} - \dfrac{1}{3*2} \int_{0}^{1}[2x-\dfrac{2x}{1+x^2}]\,dx$ $=\dfrac{\pi}{12} - \dfrac{1}{6} [x^2- ln \,(1+x^2)]_{0}^{1} =\dfrac{\pi}{12} -\dfrac{1}{6}[1-ln\, 2 - (0-ln\, 1)]=\dfrac{\pi}{12} -\dfrac{1-ln \,2}{6}$

Hence $A+B+C+D = 12+1+2+6=\boxed{21}$