How to calculate the integral of $\sum_{n=1}^\infty (1/r)^{n+1} r^2$? How to calculate this integral?
$$\int\limits_0^1 {\sum\limits_{n = 1}^\infty  {\left( {\frac{1}{r}} \right)} } ^{n + 1} r^2 dx$$
Here $r$ is a real number

I'll assume that what you wrote was intended to be written
$$
\int_0^1 \left( \sum_{n=1}^\infty \left(\frac{1}{r}\right)^{n+1}r^2 \right)dr
$$
Notice that the sum is really
$$
\sum_{n=1}^\infty \left(\frac{1}{r}\right)^{n+1}r^2=\frac{1}{r^2}r^2+\frac{1}{r^3}r^2+\frac{1}{r^4}r^2+\cdots=1+\frac{1}{r}+\frac{1}{r^2}+\frac{1}{r^3}+\cdots
$$
which is clearly a geometric series which converges if $|r|>1$ (which means we could say that we are done right here! Why?). Then we have
$$
\sum_{n=1}^\infty \left(\frac{1}{r}\right)^{n+1}r^2=\sum_{n=0}^\infty \left(\frac{1}{r}\right)^n=\frac{1}{1-\frac{1}{r}}=\frac{r}{r-1}
$$
Therefore, we have
$$
\int_0^1 \left( \sum_{n=1}^\infty \left(\frac{1}{r}\right)^{n+1}r^2 \right)dr=\int_0^1 \frac{r}{r-1} dr
$$
which clearly does not converge over $[0,1]$ as you can easily check. 
If you meant
$$
\int_0^1 \left( \sum_{n=1}^\infty \left(\frac{1}{r}\right)^{n+1}\right) r^2 dr
$$
this doesn't matter as 
$$
\int_0^1 \left( \sum_{n=1}^\infty \left(\frac{1}{r}\right)^{n+1}r^2 \right)dr=\int_0^1 \left( \sum_{n=1}^\infty \left(\frac{1}{r}\right)^{n+1} \right)r^2 dr
$$
