How do you find the indefinite integral of ∫ e^√x * arctge^√x dx / 2√x ?

1 Answer
Apr 14, 2018

#int(e^sqrtxarctan(e^sqrtx))/(2sqrtx)dx=e^(sqrtx)arctane^(sqrtx)-1/2ln(1+e^(2sqrtx))+C#

Explanation:

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#