<s>[INST] Expectation of $\frac{X_1^4}{(X_1^2 + \cdots + X_d^2)^2}$ 
Let $X_1$, $X_2$, $\cdots$, $X_d \sim \mathcal{N}(0, 1)$ and be independent. What is the expectation of $\frac{X_1^4}{(X_1^2 + \cdots + X_d^2)^2}$?

It is easy to find $\mathbb{E}\left(\frac{X_1^2}{X_1^2 + \cdots + X_d^2}\right) = \frac{1}{d}$ by symmetry. But I do not know how to find the expectation of $\frac{X_1^4}{(X_1^2 + \cdots + X_d^2)^2}$. Could you please provide some hints?
What I have obtained so far
I wanted to find $\mathbb{E}\left(\frac{X_1^4}{(X_1^2 + \cdots + X_d^2)^2}\right)$ by symmetry. But this case is different from that for $\mathbb{E}\left(\frac{X_1^2}{X_1^2 + \cdots + X_d^2}\right)$ because $\mathbb{E}\left(\frac{X_i^4}{(X_1^2 + \cdots + X_d^2)^2}\right)$ may be not equal to $\mathbb{E}\left(\frac{X_i^2X_j^2}{(X_1^2 + \cdots + X_d^2)^2}\right)$. So I need some other ideas to find the expectation.
Where this question comes from
A question in mathematics stack exchange asks for the variance of $\|Ax\|_2^2$ for a unit uniform random vector $x$ on $S^{d-1}$. My derivation shows that the answer depends sorely on the values of $\mathbb{E}\left(\frac{X_i^4}{(X_1^2 + \cdots + X_d^2)^2}\right)$ and $\mathbb{E}\left(\frac{X_i^2X_j^2}{(X_1^2 + \cdots + X_d^2)^2}\right)$ for $i \neq j$. Since
$$
\sum_{i \neq j}\mathbb{E} \left( \frac{X_i^2X_j^2}{(X_1^2 + \cdots + X_d^2)^2}\right) + \sum_i \mathbb{E}\left(\frac{X_i^4}{(X_1^2 + \cdots + X_d^2)^2}\right) = 1
$$
and by symmetry, we only need to know the value of $\mathbb{E}\left(\frac{X_1^4}{(X_1^2 + \cdots + X_d^2)^2}\right)$ to obtain other expectations.
 [/INST] The distribution of $X_i^2$ is chi-square (and also a special case of gamma).
The distribution of $\frac{X_1^2}{X_1^2 + \cdots + X_d^2}$ is thereby beta.
The expectation of the square of a beta isn't difficult.
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