The <martingale product identity> and the Itô isometry for cross terms give
$$
\mathbb E\left[B_t\int_0^te^{B_s}\,dB_s\right]
=\mathbb E\int_0^te^{B_s}\,ds.
$$
Since $B_s\sim N(0,s)$, its <moment-generating function> gives $\mathbb Ee^{B_s}=e^{s/2}$. Therefore the answer is
$$
\int_0^te^{s/2}\,ds=2(e^{t/2}-1).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 202 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-202/3/b/solution