Rewrite in terms of powers, recalling that #e^(ax)=(e^x)^a#:
#int(e^x)^2/(1+(e^x)^4)dx#
Let #u=e^x#
Then, #du=e^xdx#.
We see that since #(e^x)^2=e^x*e^x, du# shows up in our integral:
#int(e^x*color(red)(e^x))/(1+(e^x)^4)color(red)dx#
While we can write the remaining #e^x# in the numerator as #u,# since we originally said #u=e^x#.
#intu/(1+u^4)du=intu/(1+(u^2)^2)du#
We can make a second substitution here.
Let #v=u^2#
Then, #dv=2udu -> 1/2dv=udu#, as #udu# shows up in the numerator of the integral
#intcolor(red)u/(1+(u^2)^2)color(red)(du)#
Factoring the #1/2# out of the integral, we get the standard integral
#1/2int(dv)/(1+v^2)=1/2arctan(v)+C#
Rewrite in terms of #u:#
#=1/2arctan(u^2)+C#
Rewrite in terms of #x:#
#=1/2arctan((e^x)^2)+C#
#=1/2arctan(e^(2x))+C#
