How do I evaluate #int(ln(3x))^2 dx#?

1 Answer
Jan 31, 2015

Using Integration by substitution I set: #ln(3x)=t# so:
#3x=e^t#
#x=e^t/3# and
#dx=1/3e^tdt#

Your integral becomes:

#intt^2*1/3e^tdt=#

Which can now be solved by parts (twice).

By parts you have:

#intf(x)*g(x)dx=F(x)*g(x)-intF(x)*g'(x)dx#

Where:

#F(x)=intf(x)dy#
#g'(x)# is the derivative of #g(x)#
We can choose:
#f(x)=e^t#
#g(x)=t^2#

The integral becomes:

#intt^2*1/3e^tdt=1/3[e^t*t^2-inte^t2tdt]=# by parts again:
#=1/3[e^t*t^2-e^t*2t+inte^t*2dt]=#
#1/3[e^t*t^2-e^t*2t+2*e^t]=#
#1/3e^t(t^2-2t+2)#
but #ln(3x)=t# so going back to #x#:
#=1/3*3x(ln^2(3x)-2ln(3x)+2)+c#
#=x(ln^2(3x)-2ln(3x)+2)+c#