I would use "integration by parts" to solve this one.
The rule goes as follows:
#int f(x) g'(x) "d"x = f(x) g(x) - int f'(x) g(x) "d"x#
So, out of the product #x * ln(2x)# you need to pick one of the factors to differentiate and one to integrate. If #ln# is one of your factors, it is usually a good bet to choose this factor to be differentiated.
So, I would pick #f(x) = ln(2x)# and #g'(x) = x#.
Now, differentiate #f(x)# and integrate #g(x)# to compute #f'(x)# and #g(x)#:
#f(x) = ln(2x) => f'(x) = 1/(2x) * 2 = 1/x#
#g'(x) = x => g(x) = 1/2 x^2#
Now, apply the rule:
#int x ln(2x) "d"x#
#= ln(2x) * 1/2 x^2 - int 1/2 x^2 * 1 / x "d"x#
#= 1/2 x^2 ln(2x) - int 1/2 x "d"x#
#= 1/2 x^2 ln(2x) - 1/2 (1/2x^2)#
#= 1/2 x^2 ln(2x) - 1/4 x^2#
I hope that this helped!
