The answer turned out to be #4arcsin(sqrt(lnx)/2) + C#. Just a note; Wolfram Alpha gives a much more complicated answer, and it's not particularly nice of a result when, for example, you are typing it in for an online homework problem.
Wolfram Alpha gives its alternative answer as:
#= (4sqrt(lnx - 4)sqrt(lnx)ln(sqrt(lnx - 4) + sqrt(lnx)))/sqrt(-(lnx - 4)lnx)#
#= (4sqrt(lnx - 4)sqrt(lnx)ln(sqrt(lnx - 4) + sqrt(lnx)))/sqrt(4lnx - ln^2x)#
which... let's face it, looks terrible! :)
I feel like trig substitution should be done somewhere here... Let's say for now...
Let:
#u = lnx#
#du = 1/xdx#
#=> int 2/(xsqrt(4u - u^2))dx#
#= 2int 1/(sqrt(4u - u^2))du#
Okay, so this looks a little better. What if we factored something out?
#= 2int 1/(sqrtusqrt(4 - u))du#
#= 2int 1/(sqrtusqrt(2^2 - (sqrtu)^2))du#
Now, what if we said...
#sqrtu = 2sintheta#
#u = 4sin^2theta#
#du = 8sinthetacosthetad theta#
#sqrt(4 - u) = sqrt(2^2 - 2^2sin^2theta) = 2costheta#
More manageable now.
#= 2int 1/(cancel(2)cancel(sintheta)cancel(2)cancel(costheta))*cancel(8)^2cancel(sinthetacostheta)d theta#
#= 4int d theta#
Huh? How... easy?
#= 4theta#
Now, we had #2sintheta = sqrtu#, so:
#theta = arcsin((sqrtu)/2)#
Thus:
#theta = arcsin(sqrt(lnx)/2)#
And so we have:
#= color(blue)(4arcsin(sqrt(lnx)/2) + C)#
That is our answer. It seems rather intuitive considering the resemblance of the original with #1/(sqrt(1-u^2))#.
Let's try to differentiate this and see if we get back to the original.
#(dy)/(dx)[arcsinu] = 1/sqrt(1-u^2)((du)/(dx))#
Thus, with #u = sqrt(lnx)/2#, we have, after some Chain Rule action...
Cancel out terms:
#4(d)/(dx)[arcsin(sqrt(lnx)/2)] = cancel(4)*1/(sqrt(1-((sqrtlnx)/2)^2))*1/cancel(2)*1/(cancel(2)sqrt(lnx))*1/x#
Shift values around:
#= 1/(xsqrt(lnx)sqrt(1-((sqrtlnx)/2)^2))#
Multiply out the square:
#= 1/(xsqrt(lnx)sqrt(1-(lnx)/4))#
Distribute into the square root:
#= 1/(xsqrt(lnx-(lnx)^2/4))#
Factor #sqrt(1/4)# out:
#= 1/(xsqrt(1/4)sqrt(4lnx-(lnx)^2))#
Shift that around:
#= sqrt4/(xsqrt(4lnx-(lnx)^2))#
Done!
#= color(green)(2/(xsqrt(4lnx-(lnx)^2)))#
