Find all parameters $\alpha$ such that $\int_{0}^{\infty} x(x^2/2 + 1)^{\alpha}$ converges 
Find all parameters $\alpha\in\mathbb{R}$ such that $\int_{0}^{\infty} x(x^2/2 + 1)^{\alpha}$ converges.

Can you help me with it? I have no idea how to start. I know some techniques, but I still have a problem.
 A: $\int\limits_0^n x (\frac{x^2}{2}+1)^\alpha dx=\frac{(\frac{n^2}{2}+1)^{\alpha +1}-1}{\alpha +1}$ for $\alpha \ne -1$ 
$\alpha =-1$: $\,\,\int\limits_0^n x (\frac{x^2}{2}+1)^{-1}dx=\ln(\frac{n^2}{2}+1)$
$n\to\infty$: $\,\,$For $n^{2(\alpha +1)}<\infty$ it's necessary that $\alpha \lt -1$.
A: If $\alpha\ge 0$ then, as $M \to \infty$,
$$
\int_{0}^M x(x^2/2 + 1)^{\alpha}dx \ge \int_{0}^Mx\:dx \to \infty
$$ the initial integral is divergent.
Let's assume $\alpha<0$, a potential issue is only near $x \to \infty$. 
As $x \to \infty$,
$$
 x(x^2/2 + 1)^{\alpha} \sim \frac{x}{(x^2/2)^{-\alpha}}=\frac1{2^\alpha \cdot x^{-2\alpha-1}}
$$ the latter integrand is convergent near $x \to \infty$ iff $-2\alpha-1>1$ that is $\alpha<-1$.

Thus the initial integral is convergent iff $\alpha<-1$.

A: Call $x^2/2 + 1 = y$ so $\text{d}y = x\text{d}x$ and your integral becomes
$$\int_1^{+\infty} y^{\alpha}\ \text{d}y$$
From this you see immediately that the integral converges iff $\alpha < -1$
A: Let $y=x^{2}/2$
$$\int\limits_{0}^{\infty} x \left(\frac{x^{2}}{2}+1\right)^{a} \mathrm{d}x
= \int\limits_{0}^{\infty} (y+1)^{a} \mathrm{d}y
= \mathrm{B}(1,-a-1)$$
By the definition of the beta function, we have $Re(-a-1) \gt 0\,$ and thus $a \lt -1$.
