We have: $$I(a) =\int_{-\infty}^{+\infty}e^{-(ax^2+2bx)}dx $$ To prove: $$I(a) = \sqrt{\frac\pi a}e^{b^2/a}$$
I tried to differencate both sides, and got this:
Left side:
$$I'(a) = -2a\int_{-\infty}^{+\infty}xe^{-(ax^2+2bx)}dx -I(a)$$
Right side:
$$I'(a) = -I(a)\frac{b^2}{a^2}-I(a)\frac{1}{2a}$$
Use $y=x+b/a$ so $I(a)=\int_{\Bbb R}e^{-ay^2+b^2/a}dy$, so the problem reduces to proving $\int_{\Bbb R}e^{-ay^2}dy=\sqrt{\frac{\pi}{a}}$. You can do the rest yourself (there are many ways to get $I(1)$, which is the crux of it).