The value of  $\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{n^3}{(n^2+k^2)(n^2+3k^2)}$ is:

$$\begin{gathered} \underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{n^3}{n^4(1+\frac{k^2}{n^2})(1+\frac{3k^2}{n^2})} \\ =\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{k=1}^n\frac{1}{(1+\frac{k^2}{n^2})(1+\frac{3k^2}{n^2})} \\ =\underset{0}{\overset{1}{\int }}\frac{dx}{3(1+x^2)(\frac{1}{3}+x^2)} \\ =\underset{0}{\overset{1}{\int }}\frac{1}{3}\times \frac{3}{2}\frac{(x^2+1)-(x^2+\frac{1}{3})}{(1+x^2)(x^2+\frac{1}{3})}dx \\ =\frac{1}{2}\underset{0}{\overset{1}{\int }}[\frac{1}{x^2+{\left(\frac{1}{\sqrt{3}}\right)}^2}-\frac{1}{1+x^2}]dx \\ =\frac{1}{2}{\left[\sqrt{3}{\tan }^{-1}\left(\sqrt{3}x\right)\right]}_0^1-\frac{1}{2}{\left({\tan }^{-1}x\right)}_0^1 \\ =\frac{\sqrt{3}}{2}\left(\frac{\pi }{3}\right)-\frac{1}{2}\left(\frac{\pi }{4}\right)=\frac{\pi }{2\sqrt{3}}-\frac{\pi }{8} \end{gathered}$$

$=\frac{13\pi }{8.(4\sqrt{3}+3)}$