Let $T$ be a random variable with $t$ distribution with $n > 2$ degrees of freedom. I want to calculate Var$(T)$. Since $\mathbb{E}(T) = 0$, it is only necessary to calculate $\mathbb{E}(T^2)$. Now, $T$ can be written as
$$ T =Z \sqrt{\frac{n}{V}}. $$
Where $Z \sim \text{normal}(0,1)$ and $V \sim \chi^2_{n}, n \geq 1$. $Z$ and $V$ are independent.
So
$$ T^2 = Z^2 \frac{n}{V} \Rightarrow \mathbb(T^2) = \mathbb{E}(Z^2 \frac{n}{V}) = n\mathbb{E}(Z^2)\mathbb{E}\Bigl(\frac{1}{V}\Bigr). $$
We know that $\mathbb{E}(Z^2) = 1$ since $Z^2 \sim \chi^2_1$, so the only thing left to do is to calculate $\mathbb{E}\Bigl(\frac{1}{V}\Bigr)$. It is supposed to be equal to $n - 2$ because Var$(T)$ = $\frac{n}{n-2}$, but I don't know how to finish the proof.
Edit:
$\mathbb{E}(T^2)$ can also be calculated by the following integral
$$ \mathbb{E}(T^2) = \frac{\Gamma\Bigl( \frac{n+1}{2} \Bigr)}{\sqrt{n\pi} \Gamma(\frac{n}{2})} \int_{-\infty}^{\infty} x^2 \Bigl(1 + \frac{x^2}{n} \Bigr)^{- \frac{n+1}{2} } dx $$
but this integral is quite complicated (even after you finish it you just end up with a bunch of Gamma functions, which I don't know how to simplify) so I wanted to do the calculation via the first method.
The easiest way is the following
$$V\sim \chi_{(n)}^2=Gamma[\frac{n}{2};\frac{1}{2}]$$
It is well known that $\frac{1}{V}\sim$ InverseGamma with mean $\frac{\frac{1}{2}}{\frac{n}{2}-1}=\frac{1}{n-2}$
...but it is not difficult also to calculate $E[\frac{1}{V}]$ with the definition, if you prefer...
$$\mathbb{E}\Bigg[\frac{1}{V}\Bigg]=\int_0^{+\infty}\frac{1}{v}\frac{(\frac{1}{2})^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}v^{\frac{n}{2}-1}e^{-\frac{v}{2}}dv=\frac{1}{2}\frac{1}{\frac{n}{2}-1}\underbrace{\int_0^{+\infty}\frac{(\frac{1}{2})^{\frac{n}{2}-1}}{\Gamma(\frac{n}{2}-1)}v^{[\frac{n}{2}-1]-1}e^{-\frac{v}{2}}dv}_{=1}=\frac{1}{n-2}$$
:)