Quick integration question

Integrate wrt x, $\displaystyle \frac{1}{4x^2 + 4x +17}dx$.

I rearranged it into the form $\displaystyle \frac{1}{(2x+1)^2 + 16}dx$.

I know this integrates to some form of arctan, and I though that it would be

$\displaystyle \frac{1}{4} arctan{(\frac{2x+1}{4})}$.

But in the book they have the answer as $\displaystyle \frac{1}{8} arctan{(\frac{2x+1}{4})}$.

Is this because you need to divide by the derivative of $\displaystyle (2x+1)$ as well, or is there something I'm missing?

Thanks