Can the integral of Brownian motion be expressed as a function of Brownian motion and time? Let $W_t$ be standard Brownian motion, and define
$$
X_t := \int_0^t W_s ~\textrm{d}s.
$$
The marginal distributions of $X_t$ are easy to write down (see here), but it doesn't seem possible to express $X_t$ as a function $f(t,W_t)$ of time and the process $W_t$ itself. 

Is it indeed impossible? And could somebody refer me to a proof?

 A: Intuitive argument: by symmetry we should have $P(X_1 \ge 0 \mid W_1 = 0) = \frac{1}{2}$.  However, $P(f(1, W_1) \ge 0 \mid W_1 = 0)$ is either $0$ or $1$ depending on whether $f(1,0) \ge 0$.
Of course this is not really a proof because I conditioned on an event of probability 0.  So let's try to use the same idea in an actual proof.
Consider the conditional probability $Y = P(X_1 \ge 0 \mid W_1)$.
Suppose $X_1 = f(1, W_1)$ where $f$ is Borel; then we have $Y = 1_B(W_1)$ a.s., where $B = \{x : f(1,x) \ge 0\}$.  In particular we have $$P(Y \in \{0,1\}) = 1. \tag{*}$$
But since $X_1, W_1$ are both linear functionals of the Gaussian process $\{W_t\}$, they are jointly Gaussian with mean 0.  Of course $W_1$ has variance 1, and as shown in your link, $X_1$ has variance $1/3$.  Also, using Fubini's theorem we have
$$\begin{align*}
E[W_1 X_1] &= E\left[ W_1 \int_0^1 W_t\,dt\right] \\ &= E\left[\int_0^1 W_1 W_t\,dt\right] \\ &= \int_0^1 E[W_1 W_t]\,dt \\ &= \int_0^1 t\,dt = \frac{1}{2}.\end{align*}$$
So if we set $Z = W_1 - 2 X_1$, then $Z$ is a normal random variable with mean 0 and variance $1/3$ which is independent of $W_1$.  We can write the event $\{X_1 \ge 0\}$ as $\{Z \le W_1\}$.  So by independence we have $$Y = P(X_1 \ge 0 \mid W_1) = P(Z \le W_1 \mid W_1) = \Phi(\sqrt{3}W_1) \quad \text{a.s.}$$
where $\Phi$ is the standard normal cdf.  In particular $0 < Y < 1$ almost surely, contradicting (*).
