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

Question

Emily Ammerman · Accepted Answer

# 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.!

Axel Alvarez · Answer

undefined To find the antiderivative of $ (1 + e^{2x})^{1/2} $, we can use the substitution method. Let's denote $ u = 1 + e^{2x} $. Then, $ du/dx = 2e^{2x} $, and $ dx = \frac{1}{2e^{2x}} du $.

Now, substitute $ u $ and $ dx $ into the integral:

$$ \int (1 + e^{2x})^{1/2} dx = \int u^{1/2} \frac{1}{2e^{2x}} du $$

Simplify the integral:

$$ \frac{1}{2} \int \frac{u^{1/2}}{e^{2x}} du $$

Now, $ e^{2x} = u - 1 $, so:

$$ dx = \frac{1}{2(u - 1)} du $$

Substitute $ dx $ into the integral:

$$ \frac{1}{2} \int \frac{u^{1/2}}{u - 1} du $$

Now, perform partial fraction decomposition:

$$ \frac{u^{1/2}}{u - 1} = \frac{A}{u - 1} + \frac{B}{\sqrt{u}} $$

After solving for $ A $ and $ B $, integrate each term separately:

$$ \frac{1}{2} \int \left( \frac{A}{u - 1} + \frac{B}{\sqrt{u}} \right) du $$

$$ = \frac{A}{2} \int \frac{1}{u - 1} du + \frac{B}{2} \int \frac{1}{\sqrt{u}} du $$

$$ = \frac{A}{2} \ln|u - 1| + \frac{B}{2} \cdot 2\sqrt{u} + C $$

Substitute $ u = 1 + e^{2x} $ back into the expression:

$$ = \frac{A}{2} \ln|1 + e^{2x} - 1| + B\sqrt{1 + e^{2x}} + C $$

$$ = \frac{A}{2} \ln|e^{2x}| + B\sqrt{1 + e^{2x}} + C $$

$$ = Ax + B\sqrt{1 + e^{2x}} + C $$

This is the antiderivative of $ (1 + e^{2x})^{1/2} $, where $ A $ and $ B $ are constants and $ C $ is the constant of integration.