I am trying to integrate $$\int_1^\infty x(1+x^2)^p dx$$
I get down to $$\frac1{2(p+1)}\left(\lim \limits_{t \to \infty}(1+t^2)^{p+1}\right)-\frac1{2(p+1)}\lim \limits_{t \to \infty}2^{p+1}$$
The solution says that the next step is $$\frac1{2(p+1)}(0)-\frac{2^{p+1}}{2(p+1)}$$
That requires $$\lim \limits_{t \to \infty}(1+t^2)^{p+1} = 0$$ but how can this be true?
$p$ is a constant, but it can be negative or positive.
The integral converges iff $p <-1$ so, when $p <-1$ what is claimed in the solution is correct. To see that the integral converges iff $p<-1$ compare it with the following improper integral $$\int_1^{\infty} (x) (x^{2p})dx.$$