Find for what $p$, $\displaystyle \int ^{\infty}_0 x^p \arctan x dx$ converges.
By parts, it's equal to: $\displaystyle \lim_{b\to \infty}\frac 1 {p+1}x^{p+1}\arctan x |^b_0- \int ^b _0\frac {x^{p+1}}{(p+1)(1+x^2)}dx$
Now it's obvious that $p<-1$ because otherwise the left part outside of the integral goes to infinity.
So now we only have left to check for what $p$ this converges: $\displaystyle \int ^b _0\frac {x^{p+1}}{(p+1)(1+x^2)}dx$.
If we apply the ratio test with $g(x)=\frac {x^{p+1}}{x^2}$ I get that the limit $\displaystyle \lim _{x\to \infty}\frac {x^2}{(p+1)(1+x^2)}=\frac 1 {p+1}$.
Now here's the problem, since it has to be that $p<-1$, then $\frac 1 {p+1}<0$ and now what does it mean for the ratio test?
Since in absolute value it isn't zero or infinity then we have equvilience between $\frac {x^{p+1}}{(p+1)(1+x^2)}$ and $g$?
Since the integrand is non-negative, it suffices to verify finiteness:
$$ \text{For which $p$ is }\int_{\mathbb R^+}x^p\tan^{-1} x\ dx<\infty\text{?} $$ Since $\tan^{-1}x=x+o(x)$, we have the bound $\tan^{-1}x\geq cx$ for $x\in [0,\epsilon)$. Thus $p$ must satisfy $$ \int_{0}^{\epsilon}x^{p+1}<\infty, $$ which means $p>-2$.
On the other hand, since $\tan^{-1}x \geq 1$ for $x>N$ we must have $$ \int_{N}^{\infty}x^p<\infty, $$ which means $p<-1$. Therefore $p\in (-2,-1)$.
To show that every $p$ in this range works, observe that $$ \int_{\mathbb R^+}x^p\tan^{-1}x\ dx\leq \int_{0}^{N}x^{p+1}+\int_{N}^{\infty}x^p\cdot \frac{\pi}{2}<\infty. $$ We have used the bounds $\tan^{-1}x\leq x$ for $x\geq 0$ and $\tan^{-1}x\leq \frac{\pi}{2}$.