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:

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

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