Why $\frac{1}{9}$ becomes $\frac{1}{3}$ in $\frac{1}{3} \int \frac{1}{s^2+1} \, ds$? I'm trying to compute: 
$$\int\frac{1}{x^2+4x+13} \; dx$$
And I'm following the steps given by W|A here:



I don't understand why the $\cfrac{1}{9}$ becomes $\cfrac{1}{3}$ when the new substitution is made. If I switch the $s$ back to what it was, I'll get $u^2/9$ and I still need to multiply it by 9 to get $u^2+9$. I don't get why I need to change it to $1/3$.

EDIT:
$$u=x+2$$
$$s=\frac{u}{3}=\frac{1}{3}\cdot u$$
$$ds=\overbrace{\left[\frac{1}{3}\right]'u+\frac{1}{3}[u]'}^{\text{Product rule}}$$
$u=x+2$ and $u'=1$, I guess the trick is here: Instead of writing $u'$ in the product rule above as $1$, you write it as $du$.
$$ds=\left[\frac{1}{3}\right]'u+\frac{1}{3}[u]'=0\cdot u+\frac{1}{3}\cdot du=\frac{1}{3} \cdot du$$
$$ds=\frac{1}{3}du$$
"Solving" for $du$, we have:
$$3 ds=du$$
And then we switch the $ds$ with $yds$, where $y$ is whatever we have as a coefficient of $ds$.

You have
$$
\frac{1}{9}\int \frac{1}{\frac{u^2}{9} + 1} \; du = \frac{1}{9}\int \frac{1}{\left(\frac{u}{3}\right)^2 + 1} \; du.
$$
With the substitution $s = \frac{u}{3}$ you get $ds = \frac{1}{3}du$ or $\color{green}{3 ds} = \color{red}{du}$. So
$$\begin{align}
\frac{1}{9}\int \frac{1}{\left(\frac{u}{3}\right)^2 + 1} \; \color{red}{du} 
&=\frac{1}{9}\int \frac{1}{s^2 + 1} \; \color{green}{3ds}\\
&= \frac{\color{green}{3}}{9}\int \frac{1}{s^2 + 1} \; \color{green}{ds}\\
&= \frac{1}{3}\int \frac{1}{s^2 + 1} \; ds
\end{align}
$$
