Consider the conditional distribution $$f(y|x)=\frac1{(a+b*x)}e^\frac{-y}{a+b*x}, (y\geq0,0\leq x\leq1)$$
Show that the conditional mean E[Y|X] is equal to $a+b*x$. [Done]
Let X in $f(y|x)$ be uniformly distributed between 0 and 1, determine the unconditional mean E[Y].
Idea so far: $f(x)=1$ for $0\leq x \leq1$, $0$ otherwise $$f(x,y)=f(y|x)*f(x)$$ Hence:$$f(x,y)=f(y|x)=\frac1{(a+b*x)}e^\frac{-y}{a+b*x}$$ To find E[Y] I need $f(y)$, the marginal distribution of y. $$f(y)=\int_0^1\frac1{(a+b*x)}e^\frac{-y}{a+b*x}dx$$ However, I cannot solve this integral and therefore have to assume my reasoning is wrong. Any ideas on how to solve this example?