solving ODE using variation of parameters

I'm preparing for my final and came across a problem in the practice set which our professor didn't post the answers to... I tried to solve this question but failed to integrate $u_1'$ and $u_2'$. Here is the ODE: $$y''+y'-2y=\ln(x)$$

First find the solution to the corresponding homogeneous equation $y\prime\prime+y\prime-2y=0$ by using characteristic equation $r^2+r-2=0$ –  Kirthi Raman May 8 '12 at 22:59
Having found two solutions of homogeneous equation, $u_1(x)=e^{-2x}$ and $u_2(x)=e^{x}$, and computed the Wronskian $W=3e^{-x}$, we end up with general solution $u=Au_1+Bu_2$ where $$A=-\int \frac{u_2}{W} \ln(x)\,dx = -\frac13 \int e^{2x}\ln x\,dx$$ $$B=\int \frac{u_1}{W} \ln(x)\,dx = -\frac13 \int e^{-2x}\ln x\,dx$$ All we can do with these is integrate by parts (differentiating $\ln$), which expresses both integrals in terms of Exponential integral $\mathrm{Ei}$. I doubt that this was professor's intent, though.