Find the moment generating function of a non-canonized gaussian at a specific point t=0

Question

I’ve been trying to solve this problem and hit an integral I’m struggling with. Is my approach the right one? how would you continue?
The Question:
Let $X \sim N_{[1,4]}$ gaussian random variable. Find the value of the moment generating function of $X$: $(M_X(t) = E(e^{tX}))$, At the point $t=1$.
My solution attempt:
$$M_X(1) = E(e^x) = \int_{\mathbb{R}} \frac{1}{2\sqrt{2\pi}} \cdot e^{-0.5 \left(\frac{x-1}{2}\right)^2+x}dx = \int_{\mathbb{R}}\frac{1}{2\sqrt{2\pi}}e^{-\frac18(x^2-2x+1-8x)}dx$$ which doesn’t simplify nicely, so I’m struggling with how to continue.

Answer 1

This is Fluex’s solution, i claim none of it.
we can complete the square: $$-0.5\left({x-1 \over 2}\right)^2 +x = -\frac18 \cdot (x^2 -2x +1 -8x) = \left({x-5 \over 2}\right)^2 +3 $$ $$\implies \int_{\mathbb{R}}\frac{1}{2\sqrt{2\pi}}e^{-\frac18(x^2-2x+1-8x)}dx = \underbrace{\int_{\mathbb{R}} f_Y(x) dx}_{E(Y^0)=1} \cdot e^3 = e^3$$ when $Y \sim \textsf{N}_{[5,4]}$.