First note that an integer is divisible by 11 if and only if the alternating sum of the digits is divisible by 11. In our case, these are the integers where $1$, $4$ and $7$ are at the odd positions. Let $S$ be the sum of all the 5-digit integers formed by $1$, $3$, $4$, $7$, $9$ and let $T$ be the sum of those which are multiples of $11$. Then
$$\begin{align*} S &= 4!(1 + 3 + 4 + 7 + 9)(1 + 10 + 100 + 1000 + 10000) \\ &= 6399936 \\ T &= 2!2!(1 + 4 + 7)(1 + 100 + 10000) + 3!(3 + 9)(10 + 1000) \\ &= 557568. \end{align*}$$
Thus the sum is $6399936 - 557568 = 5842368$.