Integration of ln(x)

Let f denote an integral sign, I will write the integrand in square brackets. We can use integration by parts to integrate ln(x), the "trick" here is to imagin ln(x) as 1 x ln(x). Integration by parts is given by f [(u)(dv/dx)]dx = uv - f [((du/dx)(v)]dx. Using this trick allows us to set dv/dx = 1 (now do it on your own from here out!), easily integrating to give us v = x. Then we just set u = ln(x) and straightforwardly differentiate to du/dv = 1/x. Inserting this into our by parts formula gives xln(x) - f [1]dx = xln(x) - x + c