$\underset{n\to \infty }{\lim }\left(\frac{n^2}{\left(n^2+1\right)\left(n+1\right)}+\frac{n^2}{\left(n^2+4\right)\left(n+2\right)}+\frac{n^2}{\left(n^2+9\right)\left(n+3\right)}+\dots +\frac{n^2}{\left(n^2+n^2\right)\left(n+n\right)}\right)$
is equal to

$\underset{n\to \infty }{\lim }\sum _{r=1}^n\frac{\frac{1}{n}}{\left(1+\frac{r^2}{n^2}\right)\left(1+\frac{r}{n}\right)}=\underset{0}{\overset{1}{\int }}\frac{dx}{\left(1+x^2\right)\left(1+x\right)}$

Let $I=\underset{0}{\overset{1}{\int }}\frac{dx}{\left(1+x^2\right)\left(1+x\right)}$

Now $x=\tan \theta ,dx={\sec }^2\theta d\theta$

$I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{d\theta }{1+\tan \theta }=\frac{1}{2}\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{2\cos \theta d\theta }{\sin \theta +\cos \theta }$

$=\frac{1}{2}\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{\left(\sin \theta +\cos \theta \right)d\theta }{\sin \theta +\cos \theta }+\frac{1}{2}\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{\left(\cos \theta -\sin \theta \right)d\theta }{\sin \theta +\cos \theta }$

$=\frac{1}{2}\times \frac{\pi }{4}+\frac{1}{2}{\left[\ln \left|\sin \theta +\cos \theta \right|\right]}_0^{\frac{\pi }{4}}$ $=\frac{\pi }{8}+\frac{1}{4}\ln 2$