Given $X$ ~ U[1,3] and $Y$ ~ exp(X)
I'm stuck on trying to find the density and don't know if I'm taking the right approach. Usually for a problem of this sort I would condition on X and use the law of total probability, knowing that the conditional density of Y given X is exponential:
$$f_y(y)= \int_1^3f_{y|x}(y|x)f_x(x)dx = \frac{1}{2}\int_1^3xe^{-xy}dx$$
This integral doesn't come out to something "nice" which concerns me. Please let me know if this is the right track.
Your Calculation is right. The integral is very easy by parts (one passage is enough to find out the primitive) but the result is nasty so it is not forbidden to leave $f_Y(y)$ in that form, without solving it analythically
$$f_Y(y)=\int_1^3\frac{x}{2}e^{-xy}dx\mathbb{1}_{(0;+\infty)}(y)$$
In fact the density results
$$f_Y(y)=\frac{(y+1)e^{-y}-(3y+1)e^{-3y}}{2y^2}\mathbb{1}_{(0;+\infty)}(y)$$
...useful like to know if the Chinese Wall's bricks are even or odd...
As per $\mathbb{E}[Y]$ is concerned, you can calulate it without having its density:
$$\mathbb{E}[Y]=\mathbb{E}[\mathbb{E}[Y|X]]=\mathbb{E}[\frac{1}{X}]=$$
$$=\int_1^3\frac{1}{2x}dx=\frac{log3}{2}$$
EDIT: further request
Also per $P(Y>1)$ is concerned Y- density is not necessary. It is enough to have the joint density and integrate it in the correct support.
$$\mathbb{P}[Y>1]=\frac{1}{2}\int_1^3dx\int_1^{+\infty}xe^{-xy}dy=$$
$$=\frac{1}{2}\int_1^3dx\Bigg[-e^{-xy}\Bigg]_{y=1}^{y=+\infty}=\frac{1}{2}\int_1^3e^{-x}dx=\frac{e^{-1}-e^{-3}}{2}\approx 15.9\%$$