What is the second step of this standard normal MGF derivation?

On this page, it shows the derivation for the MGF of a standard normal density. I don't get why the second step has a $\sqrt{2}\sigma$ which is pulled out from the integral. Can anyone help me understand this?

$$ \begin{aligned} M_X(t) &= \int \exp(tx) \frac{1}{\sigma \sqrt{2\pi}} \exp \Big( - \frac{(x - \mu)^2}{2\sigma^2} \Big) dx\\ &= \frac{1}{\sigma \sqrt{2\pi}} \int \exp \Big(tx - \frac{(x - \mu)^2}{2\sigma^2} \Big) dx \\ &= \frac{\color{red}{\sqrt{2}\sigma}}{\sigma \sqrt{2\pi}} \int \exp \Big((\sqrt{2}\sigma u + \mu)t - u^2 \Big) dx \quad\quad u = \frac{x - \mu}{\sqrt{2} \sigma} \\ \end{aligned} $$

I think I follow everything that is going on, but the red term seems to appear for no reason. I must have missed something basic. Where does it come from?

Solution 1:

Substitution:

$$u=\frac{x-\mu}{\sqrt 2 \sigma}\implies dx=\sqrt 2 \sigma du$$