Variance of an Ito integral with Gaussian-type integrand

Consider the stochastic integral $X=\int_0^2 \sqrt{s}\,\exp\!\big(\tfrac18 W_s^2\big)\,dW_s$, where $W$ is standard Brownian motion. What is $\operatorname{Var}(X)$?

$16/3$. $X$ has mean zero, and the Ito isometry gives $\mathbb{E}[X^2]=\int_0^2 s\,\mathbb{E}[\exp(W_s^2/4)]\,ds$. Since $W_s^2=sZ^2$ and $\mathbb{E}[e^{cZ^2}]=(1-2c)^{-1/2}$, the integrand is $s(1-s/2)^{-1/2}$, and substituting $u=1-s/2$ gives $4\int_0^1(u^{-1/2}-u^{1/2})du=16/3$.