So, we want to integrate #int(e^sqrtxarctan(e^sqrtx))/(2sqrtx)dx#
The thing that stands out most is the #2sqrtx# in the denominator. This can be taken care of using a substitution:
#u=sqrtx#
#du=1/2(1/sqrtx)dx=dx/(2sqrtx)#
So, this substitution is valid. Applying it, we get
#inte^uarctan(e^u)du#
A second substitution can be used:
#v=e^u#
#dv=e^udu#
We thus have
#intarctan(v)dv#
For this final integral, we'll have to use Integration By Parts.
#a=arctan(v)#
#da=(dv)/(1+v^2)#
#db=dv#
#b=v#
#ab-intbda=varctan(v)-int(vdv)/(1+v^2)#
#int(vdv)/(1+v^2)=1/2ln(1+v^2)# -- this can be done by a mental substitution, as the differential of #1+v^2# shows up in the numerator of this integral with a factor of #2.#
This yields
#intarctan(v)dv=varctan(v)-1/2ln(1+v^2)+C#
There are no absolute value bars on the logarithm as #(1+v^2)# is always positive. Absolute value bars are redundant.
Rewrite in terms of #u:#
#e^uarctan(e^u)-1/2ln(1+e^(2u))+C#
Rewrite in terms of #x:#
#int(e^sqrtxarctan(e^sqrtx))/(2sqrtx)dx=e^(sqrtx)arctane^(sqrtx)-1/2ln(1+e^(2sqrtx))+C#