How do you integrate #int ((3+lnx)^2(2-lnx))/(4x)# using substitution?

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Answer 1 · 2016-08-26T14:54:08

#18lnx +3/2(lnx)^2-4/3(lnx)^3-1/4(lnx)^4+C#

We have that:

#int((3+lnx)^2(2-lnx))/(4x)dx#

So I would say the best substitution would be:

#u = lnx#

And from this we get :#du = 1/x dx#

We can substitute this into the integral to obtain:

#int((3+u)^2(2-u))/4du#

Now expand the brackets:

#=1/4int(9+6u+u^2)(2-u)du#

#=int(18+3u-4u^2-u^3)du#

Which integrates to give us:

#=18u +3/2u^2-4/3u^3-1/4u^4+C#

Finally we can reverse the substitution:

#=18lnx +3/2(lnx)^2-4/3(lnx)^3-1/4(lnx)^4+C#