I would need to verify if this solution is fine.
Let $W_t$ be a Brownian motion and $\lambda > 0, \text{ } \lambda \in \mathbb{R}$. Calculate $\mathbb{E} \left[W_t e^{(\lambda W_t)}\right]$.
I did
$$\mathbb{E} \left[W_t e^{(\lambda W_t)}\right] = \mathbb{E} \left[e^{\left(\frac{1}{2}\lambda^2 t\right)} W_t e^{\left(\lambda W_t - \frac{1}{2}\lambda^2 t\right)}\right]$$
Now, $e^{\left(\lambda W_t - \frac{1}{2}\lambda^2 t\right)}$ is the Radon-Nikodym derivative for changing to the measure $\mathbb{Q}$ under which $\widehat{W_t} = W_t - \lambda t$ is a Brownian motion. Therefore,
\begin{align}\mathbb{E} [W_t e^{(\lambda W_t)}] &= \mathbb{E}^{\mathbb{Q}} [e^{\left(\frac{1}{2}\lambda^2 t\right)} W_t]\\ &= \mathbb{E}^{\mathbb{Q}} [e^{\left(\frac{1}{2}\lambda^2 t\right)} \widehat{W_t} + \lambda t]\\ &= e^{\left(\frac{1}{2}\lambda^2 t\right)} \mathbb{E}^{\mathbb{Q}} [\widehat{W_t} + \lambda t]\\ &= \lambda t e^{\left(\frac{1}{2}\lambda^2 t\right)}\end{align}
Is it ok? Thanks.
Although it is correct to do it this way, it's straight overkill; from my point of view. Here is a much easier solution:
Since $W_t \sim N(0,t)$, we have
$$\mathbb{E}e^{\lambda W_t} = \exp \left(\frac{1}{2} \lambda^2 t \right).$$
Differentiating both sides with respect to $\lambda$ yields
$$\mathbb{E}(W_t e^{\lambda W_t}) = \lambda t \exp \left( \frac{1}{2} \lambda^2 t \right).$$