What is $\mathbb{E}[W(s)\mathrm{e}^{W(S)}]$ where W(S) is a standard Brownian Motion?

I know that since it is a Brownian motion that $W(s)\sim{N(0,s)}$. One approach I tried, was to see that $W(s)-W(0)\sim N(0,1)$ and to let $X = W(s)$ and then solve $\mathbb{E}[X\mathrm{e}^X]$ but I end up with some awful integral. I was wondering whether there was an easier way? Or perhaps a simple way to solve this integral: $\int_{-\infty}^{\infty} \frac{1}{\sqrt{2{\pi}s}}x\mathbb{e}^{x}\mathbb{e}^{-\frac{x^2}{2s}}$

As $W(s)\sim\mathcal N(0,s)$, you have $$ \Bbb E[\exp(\lambda W(s))] = \exp(\lambda^2 s/2). $$ Differentiate with respect to $\lambda$ and then set $\lambda=1$ to obtain $$ \Bbb E[W(s)\exp(W(s))] = s\exp(s/2). $$

This integral isn't difficult to solve. Note that $$x - \frac{x^2}{2s} = -\frac{1}{2s}(x^2 - 2sx) = -\frac{1}{2s}((x - s)^2 - s^2) = -\frac{1}{2s}(x - s)^2 + \frac{s}{2}$$

Now use the substitution $t = x - s$ and you can split your itegral into two parts that are easy to calculate.

Note that this integral is equal to the derivative of the MGF of the $\mathcal{N}(0, 1)$-distribution at $1$, so the result is $\displaystyle s\exp\Big(\frac{s}{2}\Big)$.

Set $f(x)=xe^x\in \mathbb{C}^{2}(\mathbb{R})$. We have $f'(x)=(1+x)e^x$ and $f''(x)=(2+x)e^x$. By application of Ito's lemma, We have $$f(W_t)=f(W_0)+\int_{0}^{t}f'(W_s)dW_s+\frac{1}{2}\int_{0}^{t}f''(W_s)ds\tag 1$$ We know $W_0=0$, $$W_te^{W_t}=\int_{0}^{t}(1+W_s)e^{W_s}dW_s+\frac{1}{2}\int_{0}^{t}(2+W_s)e^{W_s}ds\tag 2$$ thus $$W_te^{W_t}=\int_{0}^{t}(1+W_s)e^{W_s}dW_s+\int_{0}^{t}e^{W_s}ds+\frac{1}{2}\int_{0}^{t}W_se^{W_s}ds\tag 3$$ As a result $$\mathbb{E}\left[W_te^{W_t}\right]=\underbrace{\mathbb{E}\left[\int_{0}^{t}(1+W_s)e^{W_s}dW_t\right]}_{0}+\mathbb{E}\left[\int_{0}^{t}e^{W_s}ds\right]+\frac{1}{2}\mathbb{E}\left[\int_{0}^{t}W_se^{W_s}ds\right]\tag 4$$ In other words $$\mathbb{E}\left[W_te^{W_t}\right]=\int_{0}^{t}\mathbb{E}\left[e^{W_s}\right]ds+\frac{1}{2}\int_{0}^{t}\mathbb{E}\left[W_se^{W_s}\right]ds\tag 5$$ $W_s\sim\mathcal{N}(0,s)$, so $\mathbb{E}\left[e^{W_s}\right]=e^{\frac{1}{2}s}$ and $$\mathbb{E}\left[W_te^{W_t}\right]=2\left(e^{\frac{1}{2}t}-1\right)+\frac{1}{2}\int_{0}^{t}\mathbb{E}\left[W_se^{W_s}\right]ds\tag 6$$ Set $m(t)=\mathbb{E}\left[W_te^{W_t}\right]$, therefore

$$m(t)=2\left(e^{\frac{1}{2}t}-1\right)+\frac{1}{2}\int_{0}^{t}m(s)ds\tag 7$$ take differentiate with respect to time, $$\frac{dm(t)}{dt}=e^{\frac{t}{2}}+\frac 12 m(t)\tag 8$$ now, you should solve this ODE $$m'(t)-\frac 12 m(t)=e^{\frac{t}{2}}\tag 9$$