Question

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

Answer 1

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

Explanation:

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.!