What is the integral from -2 to 2 of (3u+1)^2 du?

Here is my work so far:

((3u+1)^3)/3

((3(2)+1)^3)/3 - ((3(-2)+1)^3)/3

= 343/3 + 125/3

= 468/3

= 156

I know that is wrong but the book says it is 52. I realized if I divide my final answer again I will get it so I was thinking that I multiplied by 3 one time too many or something. That or I made an error somewhere else.

Thanks for the help in advance.