Prove that an integral is positive For any positive integer $n$, consider 
$$\int_{0}^\infty\frac{(r^2-1)r^{n+1}}{(r^2+1)^{n+3}}dr.$$
I would like to show that it is positive. I try to write it as 
$$\int_{0}^\infty\frac{(r^2-1)r^{n+1}}{(r^2+1)^{n+3}}dr=\int_{0}^\infty\frac{r^{n+1}}{(r^2+1)^{n+2}}dr-2\int_{0}^\infty\frac{r^{n+1}}{(r^2+1)^{n+3}}dr,$$
but I am not sure that it helps. 
EDIT: According to sjasonw, the integral may not be positive as I think. I would be happy to see a proof showing that it's not positive. 
 A: Let $I$ denote the integral, then we have that
$$I = \int_{0}^\infty\frac{(r^2-1)r^{n+1}}{(r^2+1)^{n+3}}dr = \underbrace{\int_{0}^\infty\frac{r^{n+3}}{(r^2+1)^{n+3}}dr}_J - \underbrace{\int_{0}^\infty\frac{r^{n+1}}{(r^2+1)^{n+3}}dr}_K$$
Consider $J$. Let $r=1/t$, we then get that
$$J = \int_{\infty}^0 \dfrac{1/t^{n+3}}{(1/t^2+1)^{n+3}} \left(- \dfrac{dt}{t^2} \right) = \int_{0}^\infty\frac{t^{n+1}}{(t^2+1)^{n+3}}dt = K$$
Hence, $$I=0$$
A: If $$\int \frac{(r^2-1)r^{n+1}}{(r^2+1)^{n+3}}dr=I$$
$$\int \frac{(r^2-1)r^{n+1}}{(r^2+1)^{n+3}}dr=\int \frac{r^{n+1}}{(r^2+1)^{n+2}}dr-2\int_{0}^\infty\frac{r^{n+1}}{(r^2+1)^{n+3}}dr$$
Now $$\int\frac{r^{n+1}}{(r^2+1)^{n+2}}dr=\frac1{(r^2+1)^{n+2}}\int r^{n+1} dr-\int\left(\frac{d\frac1{(r^2+1)^{n+2}}}{dr}\int r^{n+1} dr\right)dr$$
$$=\frac{r^{n+2}}{(n+2)(r^2+1)^{n+2}}-\int\left( \frac{-(n+2)2r}{(r^2+1)^{n+3}}\cdot\frac{r^{n+2}}{n+2} \right)dr$$
$$=\frac{r^{n+2}}{(n+2)(r^2+1)^{n+2}}+2\int\frac {r^{n+3}}{(r^2+1)^{n+3}}dr$$ (assuming $n+2\ne 0$)
So,  $$I=\int\frac{r^{n+1}}{(r^2+1)^{n+2}}dr$$
$$=\frac{r^{n+2}}{(n+2)(r^2+1)^{n+2}}+2\int\frac {r^{n+3}}{(r^2+1)^{n+3}}dr-2\int\frac{r^{n+1}}{(r^2+1)^{n+3}}dr$$
$$=\frac{r^{n+2}}{(n+2)(r^2+1)^{n+2}}+2I$$
So, $$I=-\frac{r^{n+2}}{(n+2)(r^2+1)^{n+2}}$$
Now apply the limit.
A: Mathematica gives 
$$\int\frac{\left(r^2-1\right)r^{n+1}}{\left(r^2+1\right)^{n+3}}dr=-\frac{r^{2+n} \left(1+r^2\right)^{-2-n}}{2+n}$$
