If $X$ is a random variable with cdf $F_X$ and pdf $f_X$, and $Y=ln(f_X(X))$, then the differential entropy of $X$ is $E(-ln(f_X(X)))=-\int_{range(X)}f_X(x)ln(f_X(x))dx$. I'm making a mistake in this calculation to get the right hand side.
$F_Y(y)=P(Y\leq y)=P(ln(f_X(X))\leq y))=P(X\leq f_X^{-1}(e^y))=f_X(f_X^{-1}(e^y))=e^y$
Hence, $f_y(y)=e^y$. Therefore, $E(Y)=\int_{range(Y)}yf_Y(y)dy=\int_{range(Y)}ye^ydy=\int_{range(X)}f_X(x)ln(f_X(x))\frac{f_X'(x)}{f_X(x)}dx$, which is not the differential entropy. Where is the mistake in this calculation?