Is integer factorization an NP-complete problem?
No, its not known to be NP-complete, and it would be very surprising if it were. This is because its decision version is known to be in $\text{NP} \cap \text{co-NP}$. (Decision version: Does $n$ have a prime factor $\lt k$?)

It is in NP, because a factor $p \lt k$ such that $p \mid n$ serves as a witness of a yes instance.

It is in co-NP because a prime factorization of $n$ with no factors $\lt k$ serves as a witness of a no instance. Prime factorizations are unique, and can be verified in polynomial time because testing for primality is in P.