Finding the marginal distribution

With $\displaystyle f_X(x)\propto \frac{f_{(X|Y)}(x|y_o)}{f_{(Y|X)}(y_o|x)}$

I am given that

$\displaystyle f_{(X|Y)}(x|y)\propto (y)exp(-yx)$ at interval $\displaystyle 0\le x<b<+\infty$

and

$\displaystyle f_{Y|X}(y|x)\propto \frac{(x)exp(-xy)}{\int_0^b (x)exp(-xy)dy}$ at interval $\displaystyle 0\le y<b<+\infty$

Let $\displaystyle y_o=b/2$ and $\displaystyle \int_0^b (x)exp(-xy)dy=1-exp(-bx)$

$\displaystyle \frac{f_{(X|Y)}(x|y_o)}{f_{Y|X}(y_o|x)}=(b)exp(-bx)/\frac{(x)exp(-bx)}{1-exp(-bx)}=\frac{b}{x}(1-exp(-bx))$

Now In order to find the normalising constant c such that $\displaystyle f_X(x)=(c)\frac{f_{(X|Y)}(x|y_o)}{f_{Y|X}(y_o|x)}$, I want to integrate the right hand side.. i.e $\displaystyle \displaystyle\int^b_0\frac{1}{x}(1-exp(-bx))$ but it seems can't be done? For example: $\displaystyle \int_0^b\frac{1}{x}$ undefined. May I know where the problems lie at?