Limit + Summation = Integration

By Riemann Sums we have: $\begin{aligned} S &= \lim_{n\to\infty}\sum_{k=1}^{n}\dfrac{n^2}{n^3+k^3}\\ &= \lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{1+\left(\frac{k}{n}\right)^3}\\ &= \int_{0}^{1} \dfrac{\mathrm{d}x}{x^3+1}\\ &= \dfrac{1}{3}\int_{0}^{1} \left(\dfrac{1}{x+1}-\dfrac{x-2}{x^2-x+1}\right) \mathrm{d}x\\ &= \dfrac{1}{3}\int_{0}^{1} \dfrac{\mathrm{d}x}{x+1} - \dfrac{1}{3} \int_{0}^{1} \dfrac{x-2}{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} \mathrm{d}x \end{aligned}$ For the second integral make the change $x-\frac{1}{2}=\frac{\sqrt{3}}{2}\tan \theta$ , then: $\begin{aligned} S &= \dfrac{1}{3}\ln(x+1)\Big|_{x_1=0}^{x_2=1} + \dfrac{1}{3} \int\limits_{-\pi/6}^{\pi/6} (\sqrt{3}-\tan \theta)\mathrm{d}\theta\\ &= \dfrac{1}{3}\ln(2) + \dfrac{1}{3} \left(\sqrt{3}\theta+\ln(\cos \theta)\right) \Bigr|_{\theta_1=-\pi/6}^{\theta_2=\pi/6}\\ &= \dfrac{1}{3}\ln(2) + \dfrac{1}{3} \left(\dfrac{\sqrt{3}\pi}{3}\right)\\ &= \dfrac{3\ln(2) + \sqrt{3}\pi}{9}\\ &= \dfrac{\sqrt{3}\pi + \ln(8)}{9} \end{aligned}$ So, the answer is $3+8+9=\boxed{20}$ .