Let f(x)=xln(x)-x. Find f'(x). Hence or otherwise, evaluate the integral of ln(x^3) between 1 and e.

We use the product rule with u=x and v=ln(x) (so u'=1 andv'=1/x) to differentiate xln(x) to ln(x)+1, and -x just differentiates to -1, hence we have. f'(x)=ln(x).
Now note that ln(x^3)=3ln(x) using properties of logarithms.Hence, we are just integrating 3ln(x). We know, from the first part, that ln(x) will integrate to xln(x)-x, and we require 3 lots of this, so 3ln(x) integrates to 3xln(x)-3x. Plugging in 1 and e into this formula, we get that the integral of ln(x^3) between 1 and e is (3eln(e)-3e)-(3ln(1)-3) = (3e-3e)-(30-3) = -3 (since ln(e)=1 and ln(1)=0).