Can you show me why the integral of 1/x is the natural log of x?

We can break this down into steps, going deeper each time. First we might just say: well, since integration is fundamentally the inverse process of differentiation and we know that the derivative of ln(x) is 1/x, then the integral of 1/x must be ln(x).
But hold on... how do we know that d/dx(ln x) = 1/x? Well, looking at ln(x), we cannot differentiate it directly but we do know that its inverse is e^x. Imagining a plot of the function ln(x), we know that inverting it is equivalent to switching round the axes or reflecting in y = x. From this, it is easy to see graphically that the derivative of the inverse function is the inverse of the derivative of the function. So if y = ln(x) : d/dx(ln x) = 1/ d/dy(e^y). However, the derivative of e^y is e^y from the definition of the exponential function and so this simplifies to: d/dx(ln x) = 1/ e^y = 1/x, showing that the integral of 1/x is ln(x).