$u(t,x)$ is a solution of the Schrödinger equation $iu_{t}+\Delta u=0.$
$E(t) := \int |u(t,x)|^2 dx$ is conserved and since $u\bar{u}=|u|^2$ then $$\frac{d}{dt}E(t)=\int (\bar{u}u_{t}+\bar{u_{t}} u)\, dx.$$ Now using the Schrodinger equation $iu_{t}+\Delta u=0,$ we see that $u_{t}=i\Delta u$. Substituting this into $\frac{d}{dt}E(t)$ we get $$\frac{d}{dt}E(t)=\int[\bar{u}(i\Delta u)+\overline{(i\Delta u)}u]\,dx = 2i\operatorname{Im}\int|\nabla u|^{2}dx.$$
I am unsure with how to calculate this integral and give the required result, I know it involves integration by parts but can anyone show me how to actually perform this calculation?
I think you made a slight mistake in the the signs. Anyway, here is my computation: \begin{align} \frac{d}{dt}E(t)&=\int (\bar{u}u_{t}+\bar{u_{t}} u)\, dx \\ &=-\int[\bar{u}(i\Delta u)+\overline{(i\Delta u)}u]\,dx \\ &=-i\int[\bar{u} (\Delta u)-(\Delta\overline{u})u]\,dx \end{align}
Note the relative minus sign between the two terms. Now you partial integration to transform $\bar{u} (\Delta u)$ into $(\Delta\overline{u})u$, and you see that both halves just cancel.
On second thought, your calculation is fine too. Your last expression is "imaginary part of something real", which is zero.