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Answer 1 · 2017-06-07T21:02:27

#1/2xlnx - x/2 + C#

This should be a familiar integral if you simply recognize that #lnsqrtx = lnx^"1/2" = 1/2lnx#, and if you've done #int lnxdx# before.

Therefore:

#int lnsqrtxdx#

#= 1/2 int lnxdx#

Now, for this, use integration by parts. Have #u# be what you can differentiate and #dv# be what you can easily integrate.

Let:

#u = lnx#
#du = 1/xdx#
#dv = dx#
#v = x#

Thus:

#1/2 int udv = 1/2 [uv - intvdu]#

#= 1/2 int lnxdx#

#= 1/2[xlnx - int x cdot 1/xdx]#

#= color(blue)(1/2xlnx - x/2 + C)#