How do you find the antiderivative of #(1 + e^(2x)) ^(1/2)#?

Question

1345 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-08T11:54:22

# 1/2ln|(1-sqrt(1+e^(2x)))/(1+sqrt(1+e^(2x)))|+sqrt(1+e^(2x))+C.#

Let us subst. #e^x=tany rArr e^xdx=sec^2ydy, or, dx=sec^2y/e^x*dy=sec^2y/tany*dy=1/(cosysiny)dy#

#:. I=intsqrt(1+e^(2x))dx#

#=int{sqrt(1+tan^2y)/(cosysiny)}dy,#

#=int1/(cos^2ysiny)dy=int{(siny)/(cos^2ysin^2y)}dy.#

Hence, #cosy=t rArr -sinydy=dt, and, :.,#

#I=-int1/{t^2(1-t^2)}dt,#

#=int1/{t^2(t^2-1)}dt=int[{t^2-(t^2-1)}/{t^2(t^2-1)}]dt,#

#=int[t^2/{t^2(t^2-1)}-(t^2-1)/{t^2(t^2-1)}]dt,#

#=int(1/(t^2-1)-1/t^2)dt,#

#=1/2ln|(t-1)/(t+1)|+1/t,#

#=1/2ln|(cosy-1)/(cosy+1)|+1/cosy,#

#=1/2ln|(1-secy)/(1+secy)|+secy.#

Since, #tany=e^x rArr secy=sqrt(1+tan^2y)=sqrt(1+e^(2x)),# we get,

#I=1/2ln|(1-sqrt(1+e^(2x)))/(1+sqrt(1+e^(2x)))|+sqrt(1+e^(2x))+C.#

Enjoy Maths.!