Problem and Answer:
$$\mathbb{E} [X_t] = \int^\infty_{-\infty} e^{\mu t + \sigma x}\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}dx= e^{\mu t + t\frac{\sigma^2}{2}}$$
I don't understand the middle steps to get to the answer. I get that $ut$ is constant, and you bring $\left(\frac{1}{\sqrt{2\pi t}}\right)$ in front of the integral, but I'm not understanding what I should do after that.
$$I = \int_{-\infty}^\infty e^{\mu t+\sigma x} \frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}dx = \frac{1}{\sqrt{2\pi t}} e^{\mu t} \int_{-\infty}^\infty e^{\sigma x -\frac{x^2}{2t}}dx$$ We now need to complete the square of the exponent, $$\sigma x -\frac{x^2}{2t} = -\frac{1}{2t}(x^2-2t\sigma x) = -\frac{1}{2t}(x^2-2t\sigma x + (t\sigma)^2) + \frac{(t\sigma)^2}{2t} = -\frac{1}{2t}(x - t\sigma)^2 + \frac{1}{2}t\sigma^2$$ Now, let $u= x-t\sigma$ and $du = dx$.
$$I = \frac{1}{\sqrt{2\pi t}} e^{\mu t + \frac{1}{2}t\sigma^2} \int_{-\infty}^\infty e^{-\frac{1}{2t}u^2}dx = \frac{1}{\sqrt{2\pi t}} e^{\mu t + \frac{1}{2}t\sigma^2} \sqrt{2\pi t} = e^{\mu t + \frac{1}{2}t\sigma^2}$$
Note that I used the fact that $\int_{-\infty}^\infty e^{-ax^2} = \sqrt{\frac{\pi}{a}}$