Is log(n) equivalent to (log(n))^x for big-O analysis?

Question

My professor noted that we could treat any logarithmic function with an exponent as equivalent to log(n) for the purposes of big-O analysis.

ie. $(n log(n) + 1)^2 + (log(n) + 1)(n^2 + 1)$

From the left I would get $(n^2)(log(n))^2$ and from the right I would get $(n^2)log(n)$. According to my professor I can just say this function is $O(n^2log(n))$. However, I don't see how these functions grow in the same order. Looking at their graphs it would seem they are quite different.