Hello, It has been over a year since I last took calculus. And I don't recall how to take the integral of a natural logarithmic function. Here is the question that I am supposed to integrate. double integral 1/(x+y) dA where R = [1,2] X [0,1] So what I did first was integrate with respect to y first. I ended up with ln(x+y) with an upper limit of 1 and a lower limit of 0. Once simplified I get ln(x+1) - ln x or ln( (x+1)/x ) Now I have to integrate with respect to x. But I can't remember how to take the integral of a natural log function. How do I proceed from here? I can't remember if I can do the following: Let G(x) = integral of ln( (x+1)/x ) dx then e^G(x) = integral of e^ln( (x+1)/x ) dx which would simplify to integral of (x+1)/x dx. After I get a solution to the above equation I would then take the log of ln e^G(x) = ln (answer) to get G(x) = ln (answer). Can I do that? I can't remember. If not, how do I proceed from here? Any help is appreciated. Thankyou.