Probabilty Density for $x$

Suppose that I have the function $$x(t)=e^{-b(W(t))^2} (1)$$ where $W(t)$ is just a Wiener process (i.e. a Gaussian in general). I want to know what the probability density for $x$, $P(x)$, is. I started off by just assuming that I want to measure the expectation value of an observable $f(x)$, so $$<f(x)>=int_{W=0}^{W=t}{P(W)f(g(W))dW} , x=g(W) $$ Then I transformed variables from $W$ to $t$ and I got $$<f(x)>=int_{x=g(0)}^{x=g(t)}P(g^{-1}(x))f(x)(frac{dW}{dx})dx=int_{x=g(0)}^{x=g(t)}frac{P(g^{-1}(x))}{g'(g^{-1}(x))}f(x)dx$$ so I just assume that the probability for x is $$P(x)=frac{P(g^{-1}(x))}{g'(g^{-1}(x))}$$ Since $x=g(W)$, then $W=g^{-1}(x)$, but from (1) I get $$frac{dW}{dx}=frac{1}{2xsqrt{blnx}}$$ and assuming that $sigma^2=V=t$, I get $$P(x)=frac{e^{lnx/{2bt}}2sqrt{blnx}}{xsqrt{2{pi}t}}$$ Is this right? Shouldn’t I get a Gaussian? Also, was I right to take the values of $x$ to be $[0,t]$ and not $(-infty,+infty)$?