<s>[INST] Limit of a Riemann sum: $\lim_{n\to\infty} {n^5 \sum^n_{r=0}\frac1{(n^2+r^2)^3}} $ Required to find 
$\lim_{n\to{\infty}} {n^5 \sum^n_{r=0}\frac{1}{(n^2+r^2)^3}} $
$\lim_{n\to{\infty}} \frac{1}{n} \sum^n_{r=0}(\frac{n^2}{n^2+r^2})^3$
$\lim_{n\to{\infty}} \frac{1}{n} \sum^n_{r=0}(\frac{1}{1+\frac{r^2}{n^2}})^3$
$\lim_{n\to{\infty}} \frac{1-0}{n} \sum^n_{r=0}f[0+r(\frac{1-0}{n})]$, where $f(x)=(\frac{1}{1+x^2})^3$.
Since $f(x)$ is continuous on $]0,1[$ 
$\frac{1}{n} \sum_{r=1}^n f(\frac{r}{n})$ is a Riemann sum.
$\lim_{n\to{\infty}}\frac{1}{n} \sum_{r=1}^n f(\frac{r}{n})=\int_0^1 \frac{1}{(1+x^2)^3}$ $dx$
I am getting problems to compute this integral. 
I used a substitution of $x=\tan{u}$
My integral becomes $\int^\frac{\pi}{4}_0 (\cos{u}-\sin^2{u}\cos{u})du=\frac{5\sqrt{2}}{2}$
Hence $\lim_{n\to{\infty}}\frac{1}{n} \sum_{r=1}^n f(\frac{r}{n})=\int_0^1 \frac{1}{(1+x^2)^3}$ $dx=\frac{5\sqrt{2}}{2}$
But the limit is defined to be $\frac{3\pi}{32}+\frac{1}{4}$. I am getting it to be different.
 [/INST] $$n^5\sum_{k=0}^n\frac1{(n^2+k^2)^3}=\frac1n\sum_{k=0}^n\frac1{\left(1+\left(\frac kn\right)^2\right)^3}\xrightarrow[n\to\infty]{}\int_0^1\frac{dx}{(1+x^2)^3}$$
Your substitution is fine...yet you must also change the limits accordingly:
$$x=\tan u\implies\begin{cases}x=0=\tan 0\implies u=0\\{}\\x=1=\tan u\implies u=\frac\pi4\end{cases}\;\;,\;\;\;dx=\frac{du}{\cos^2u}$$
and our integral becomes, using $\;1+\tan^2u=\frac1{\cos^2u}\;$ :
$$\int_0^{\pi/4}\frac{du}{\cos^2u}\cdot\frac1{(1+\tan^2u)^3}=\int_0^{\pi/4}\cos^4u\;du=\left.\frac1{32}\left(12u+8\sin2u+\sin4u\right)\right|_0^{\pi/4}=$$
$$=\frac1{32}\left(3\pi+8\right)=\frac{3\pi}{32}+\frac14$$
</s>