Compute $\mathbb{E}\big[\exp(XY+X)\big]$ where $X, Y$ are independent uniformly distributed over $[0,1]$ r.v's. 
Compute $\mathbb{E}\big[\exp(XY+X)\big]$ where $X, Y$ are independent uniformly distributed over $[0,1]$ r.v's.

I first computed $\mathbb{E}\big[\exp(XY+X)\mid X\big]$.
Because $X$ and $Y$ are independent then  $$\mathbb{E}\big[\exp(XY+X)\mid X\big] = \phi(X)\,,$$
where $$\phi(x) =\mathbb{E}\big[\exp(xY+x)\big] = \int_0^1 \exp(xy+x)dy = \frac{\exp x(\exp x - 1)}{x}.$$
so  $$\mathbb{E}\big[\exp(XY+X)\big] =  \int_0^1 \frac{\exp x(\exp x - 1)}{x} dx\,,$$
which I don't know how to compute. 
Could someone check if I'm good till now and maybe help me continue? Thanks!
 A: You have done great so far.  But I dont think there is an elementary expression for the last integral.  But you can note that the exponential integral $\operatorname{Ei}$ is given by
$$\operatorname{Ei}(t)=\mathrel{-\!\!\!\!\!\!\;\!\!\int}_{-\infty}^{t}\frac{e^{s}}{s}ds.$$
(Here $\mathrel{-\!\!\!\!\!\;\!\!\int}$ is the Cauchy principal value integral.)
In other words,
$$\int\frac{e^{s}}{s}ds=\operatorname{Ei}(s)+C.$$
This also shows that
$$\int\frac{e^{ks}}{s}ds=\operatorname{Ei}(ks)+C.$$
The required integral is
\begin{align}I&=\int_0^1\frac{e^x(e^x-1)}{x}dx=\lim_{\epsilon\searrow0}\int_{\epsilon}^1\left(\frac{e^{2x}}{x}dx-\int_\epsilon^1\frac{e^x}{x}dx\right)\\
&=\lim_{\epsilon\searrow0}\Big(\big(\operatorname{Ei}(2)-\operatorname{Ei}(2\epsilon)\big)-\big(\operatorname{Ei}(1)-\operatorname{Ei}(\epsilon)\big)\Big)\\
&=\operatorname{Ei}(2)-\operatorname{Ei}(1)-\lim_{\epsilon\searrow0}\big(\operatorname{Ei}(2\epsilon)-\operatorname{Ei}(\epsilon)\big).\end{align}
Now
$$\operatorname{Ei}(2\epsilon)-\operatorname{Ei}(\epsilon)=\int_{\epsilon}^{2\epsilon}\frac{e^s}{s}ds=\int_\epsilon^{2\epsilon}\frac{1+O(\epsilon)}{s}ds=\int_{\epsilon}^{2\epsilon}\frac{ds}{s}+O(\epsilon)=\ln 2+O(\epsilon).$$
This gives
$$I=\operatorname{Ei}(2)-\operatorname{Ei}(1)-\ln2\approx 2.366.$$
A: You can't get an explicit answer.  However let $u=\exp(x)$ then $du=\exp(x)dx$ and your integral becomes $\int_1^e \frac{u-1}{\ln(u)}du=\text{li}(e^2)-\text{li}(e)=\int_e^{e^2}\frac{1}{\ln(u)}du$.  Unfortunately there is no explicit formula for $\text{li}(u)$.
I calculated this result and got I=3.0591165.
In addition, due to the singularity of the integrand at $u=1$, it is necessary to subtract $\lim_{\epsilon\to 0}\int_{(1+\epsilon)}^{(1+\epsilon)^2}\frac{1}{ln(u)}du=ln(2)=0.693147181$
Net result$=2.365969319$.
